\(Q=\frac{1+\text{ax}}{1-\text{ax}}\sqrt{\frac{1-bx}{1+bx}}\)

Ta có: \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow\text{ax}=\sqrt{\frac{2a-b}{b}}\Rightarrow1+\text{ax}=1+\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}}\)

\(1-\text{ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)

\(\Rightarrow\frac{1+\text{ax}}{1-\text{ax}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}-\sqrt{2a-b}}=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2b-2a}\left(1\right)\)

 \(bx=\frac{b}{a}\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}\left(2a-b\right)}{a}\Rightarrow\hept{\begin{cases}1-bx=\frac{a-\sqrt{b\left(2a-b\right)}}{a}\\1+bx=\frac{a+\sqrt{b\left(2a-b\right)}}{a}\end{cases}}\)

\(\Rightarrow\frac{1-bx}{1+bx}=\frac{a-\sqrt{b\left(2a-b\right)}}{a+\sqrt{b\left(2a-b\right)}}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{a^2-2ab+b^2}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{\left(a-b\right)^2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow Q=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2\left(b-a\right)}.\frac{a-\sqrt{b\left(2a-b\right)}}{a-b}=\frac{\text{[}2a+2\sqrt{b\left(2a-b\right)}\text{]}\left(a-b\sqrt{2a-b}\right)}{2\left(a-b\right)^2}\)

\(\Rightarrow\frac{2\left[a^2-b\left(2a-b\right)\right]}{2\left(a-b\right)^2}=\frac{2\left(a^2-2ab+b^2\right)}{a\left(a-b\right)^2}=1\)

ta có \(A=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}>0\forall x>0;x\ne1\left(1\right)\)

\(A-2=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}-2=\frac{\sqrt{x}+2-2\sqrt{x}-2x-2}{\sqrt{x}+x+1}=\frac{-\sqrt{x}-2x}{\sqrt{x}+x+1}\le0\forall x\)

=> A =< 2 (2)

Từ (1) và (2) => 0<A =< 2 => A={1;2}

với A=1 \(\Rightarrow\frac{\sqrt{x}+2}{x+\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}+2=x+\sqrt{x}+1\Leftrightarrow x=1\left(ktm\right)\)

với A=2\(\Rightarrow\frac{\sqrt{x}+2}{x+\sqrt{x}+1}=2\Leftrightarrow\sqrt{x}+2=2\sqrt{x}+2x+2\)

\(\Leftrightarrow2x+\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=0\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(ktm\right)\)

Vậy không có giá trị x để A là số nguyên

\(\left(\sqrt{6}-2\sqrt{3}+5\sqrt{2}-\frac{1}{2\sqrt{8}}\right).2\sqrt{6}\)

\(=2.6-12\sqrt{3}+20\sqrt{3}-\frac{\sqrt{3}}{2}\)

\(=\frac{24+15\sqrt{3}}{2}\)

\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)

\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}=14\sqrt{3}\)

câu tiếp tương tự câu thứ 2 nha

Cho hàm số bậc nhất y = ax + 7

Thế x = 7 , y = 8 vào hàm số ta được :

8 = a.7 + 7

<=> 8 = 7a + 7

<=> 7a = 1

<=> a = 1/7

Vậy a = 1/7 

cho hàm số bậc nhất y= ax + 7 

thế x=7, y=8 vào hàm số ta được :

<=> 8=7a + 7

<=> 7a=1

<=> a=1/7 

vậy a=1/7   @_@ cảm ơn bạn 

a,\(\sqrt{1-x}=\sqrt[3]{27}\left(đk:x\le1\right)\Leftrightarrow\sqrt{1-x}=3\)

\(< =>\sqrt{1-x}^2=9< =>1-x=9< =>x=-8\)tm

b,\(\sqrt{x^2-10x+25}=x+1\)

\(< =>\sqrt{\left(x-5\right)^2}=x+1\)

\(< =>|x-5|=x+1\)

\(< =>\orbr{\begin{cases}-x+5=x+1\left(x< 5\right)\\x-5=x+1\left(x\ge5\right)\end{cases}}\)

\(< =>\orbr{\begin{cases}2x=4< =>x=2\left(tm\right)\\-5-1=0\left(vo-li\right)\end{cases}}\)

c, Đặt \(\sqrt{x}=t\left(t\ge0\right)\)khi đó pt tương đương

\(t^2+t-6=0< =>t^2-2t+3t-6=0\)

<\(< =>t\left(t-2\right)+3\left(t-2\right)=0< =>\left(t+3\right)\left(t-2\right)=0\)

\(< =>\orbr{\begin{cases}t+3=0\\t-2=0\end{cases}}< =>\orbr{\begin{cases}t=-3\left(ktm\right)\\t=2\left(tm\right)\end{cases}}\)

khi đó ta được \(\sqrt{x}=t< =>x=4\)

a) \(\sqrt{1-x}=\sqrt[3]{27}\)

\(\Leftrightarrow\sqrt{1-x}=3\)

\(\Leftrightarrow1-x=9\)

\(\Rightarrow x=-8\)

b) \(\sqrt{x^2-10x+25}=x+1\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+1\)

\(\Leftrightarrow\left|x-5\right|=x+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=x+1\\x-5=-x-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}0=6\left(vl\right)\\2x=4\end{cases}}\Rightarrow x=2\)

c) \(x+\sqrt{x}-6=0\)

\(\Leftrightarrow\left(x+3\sqrt{x}\right)-\left(2\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=-3\left(vl\right)\end{cases}}\Rightarrow x=4\)

\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)

\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)

\(\Leftrightarrow C=-3\)

+) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)

\(=3\sqrt{4.5}-2\sqrt{9.5}+4\sqrt{5}\)

\(=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)

\(=4\sqrt{5}\)

+) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=\left(2\sqrt{7}-\sqrt{28}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=\left(2\sqrt{7}-2\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=7+7\sqrt{8}\)

dell trả lời