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\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
a: Thay x=49 vào A, ta được:
\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)
b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)
P>1 khi P-1>0
=>\(\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}>0\)
=>\(\sqrt{x}-2>0\)
=>\(\sqrt{x}>2\)
=>x>4
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)
Tính giá trị biểu thức
A= \(\left(4x^5+4x^4-5x^3+2x-2\right)^2+2020\) khi \(x=\dfrac{\sqrt{5}-1}{2}\)
Lời giải:
$x=\frac{\sqrt{5}-1}{2}$
$2x=\sqrt{5}-1$
$2x+1=\sqrt{5}\Rightarrow (2x+1)^2=5$
$\Leftrightarrow 4x^2+4x-4=0$
$\Leftrightarrow x^2+x-1=0$
Khi đó:
\((4x^5+4x^4-5x^3+2x-2)^2\)
\(=[4x^3(x^2+x-1)-x^3+2x-2]^2\)
\(=(-x^3+2x-2)^2=[-x(x^2+x+1)+(x^2+x-1)-1]^2\)
\(=(-1)^2=1\)
\(a,\sqrt{x+1}< 2\Leftrightarrow\left\{{}\begin{matrix}x+1< 4\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x\ge-1\end{matrix}\right.\\ \Leftrightarrow-1\le x< 3\)
\(d,\sqrt{2x+1}\ge3\Leftrightarrow2x+1\ge9\Leftrightarrow x\ge4\)
a, \(A=\dfrac{\sqrt{x}-1}{x^2-x}:\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)=\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}\pm1\right)}:\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{1}{x\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}=1\)
b, Cho A = 1 rồi còn gì, hay đề lỗi bạn ?
\(x=4+2\sqrt{3}=\sqrt{3}^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
xem là bài mình làm có sai đâu ko nhé nếu rút gọn ra kq khác thì thay bên trên vào nhé
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
\(P=\dfrac{3\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}=3+\dfrac{4}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(4\right)=\left\{1;2;4\right\}\left(\sqrt{x}-2\ge2-2=0\right)\\ \Leftrightarrow\sqrt{x}\in\left\{3;4;6\right\}\\ \Leftrightarrow x\in\left\{9;16;36\right\}\left(tm\right)\)
Vậy chọn D
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
a) \(x^2=49\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)