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a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
a)
Đặt
\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)
Khi đó:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)
\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)
\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)
\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)
Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)
Ta có:
Áp dụng tính chất dãy tỉ số bằng nhau thì:
\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)
Khi đó:
\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)
Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)
Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix}
1\\
-5\end{matrix}\right.\)
\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
Mk sửa lại đề nha
\(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\left(ĐKXĐ:x\ne25\right)\)
\(A=\left(\frac{x-5\sqrt{x}-x+25}{x-25}\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(A=\left(\frac{25-5\sqrt{x}}{x-25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(A=\left(\frac{5.\left(5-\sqrt{x}\right)}{x-25}\right):\left(\frac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
