M=\(\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\left(1+\dfrac{1}{a}\right)\)
CM rằng giá trị m ko phụ thuộc vào a
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NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1
\(A=\dfrac{2}{\sqrt{ab}}:\left(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}}\right)^2-\dfrac{a+b}{\left(\sqrt{a}-b\right)^2}\)
\(=\dfrac{2}{\sqrt{ab}}.\dfrac{ab}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\dfrac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\dfrac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\dfrac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(==\dfrac{-\left(a-2\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\dfrac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
AH
Akai Haruma
Giáo viên
1 tháng 4 2021
Lời giải:
ĐK: $x>0; a\neq 1; a\neq 4$
a)
$M=\frac{\sqrt{a}-(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}-1)}:\frac{(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}-2)(\sqrt{a}+2)}{(\sqrt{a}-2)(\sqrt{a}-1)}$
$=\frac{1}{\sqrt{a}(\sqrt{a}-1)}:\frac{3}{(\sqrt{a}-2)(\sqrt{a}-1)}=\frac{1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}-2)(\sqrt{a}-1)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}$
b)
$M>\frac{-1}{2}\Leftrightarrow \frac{\sqrt{a}-2}{3\sqrt{a}}+\frac{1}{2}>0$
$\Leftrightarrow \frac{5\sqrt{a}-4}{6\sqrt{a}}>0$
$\Leftrightarrow 5\sqrt{a}-4>0$
$\Leftrightarrow a>\frac{16}{25}$
Kết hợp với ĐKXĐ thì $a>\frac{16}{25}; a\neq 1; a\neq 4$
10 tháng 11 2018
\(Q=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a\right)}\right).\dfrac{a+1}{a}\)
\(Q=\dfrac{\left(1-\sqrt{a}\right)\left(1+a\right)+\left(1+\sqrt{a}\right)\left(1+a\right)-2\left(a^2+1\right)}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a\right)}.\dfrac{a+1}{a}\)
\(Q=\dfrac{\left(1+a\right)\left(1-\sqrt{a}+1+\sqrt{a}\right)-2a^2-2}{2a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(Q=\dfrac{2\left(1+a\right)-2a^2-2}{2a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(Q=\dfrac{1+a-a^2-1}{a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(Q=\dfrac{a-a^2}{a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(Q=\dfrac{a\left(1-a\right)}{a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(Q=\dfrac{a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{a\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=1\)
vậy
NQ
4 tháng 12 2017
a) Q\(=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right).\left(1+\dfrac{1}{a}\right)\) tồn tại :
\(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\2-2\sqrt{a}\ne0\\1-a^2\ne0\\a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
10 tháng 7 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}\)
\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)
c) Để M=-4 thì \(-2\sqrt{a}=-4\)
\(\Leftrightarrow\sqrt{a}=2\)
hay a=4(thỏa ĐK)
AT
30 tháng 7 2021
\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)
\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)
17 tháng 10 2023
3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{6}{3\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\)
b: M>1/3
=>M-1/3>0
=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)
=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
=>\(3-\sqrt{x}>0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ
=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ
Dấu = xảy ra khi x=0
H9
HT.Phong (9A5)
CTVHS
28 tháng 7 2023
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
15 tháng 10 2017
điều kiện xác định là : \(a>0;a\ne1\)
ta có : \(P=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right)\dfrac{\left(\sqrt{a}-1\right)\left(a-1\right)}{\sqrt{a}}\)
\(P=\left(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(P=\left(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(P=\left(\dfrac{a-\sqrt{a}+2\sqrt{a}-2-\left(a+\sqrt{a}-2\sqrt{a}-2\right)}{\sqrt{a}+1}\right)\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(P=\dfrac{a-\sqrt{a}+2\sqrt{a}-2-a-\sqrt{a}+2\sqrt{a}+2}{\sqrt{a}+1}.\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(P=\dfrac{2\sqrt{a}}{\sqrt{a}+1}.\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{2}{\sqrt{a}+1}.\sqrt{a}-1=\dfrac{2\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)
\(P=\dfrac{2\sqrt{a}-2}{\sqrt{a}+1}\) (biểu thức này luôn phụ thuộc vào biến) (đpcm)
\(M=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\left(1+\dfrac{1}{a}\right)\)
\(M=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right)\left(1+\dfrac{1}{a}\right)\)
\(M=\left(\dfrac{\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{2}{2\left(1-a\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{1}{1-a}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{1+a-a^2-1}{\left(1-a\right)\left(1+a\right)}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\dfrac{a-a^2}{a\left(1-a\right)}\)
\(M=\dfrac{a\left(1-a\right)}{a\left(1-a\right)}=1\)
--> giá trị của M ko phụ thuộc vào A
cái này mà là toán lớp 6 hả