a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

\(\Leftrightarrow\left(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}+\sqrt{a}\right).\dfrac{\sqrt{a}-2}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}-2}+\sqrt{a}.\dfrac{\sqrt{a}-2}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{1+\sqrt{a}\sqrt{a}-2}{\sqrt{a}-2}.\dfrac{\sqrt{a}-2}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{1+a-2a}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\Leftrightarrow\sqrt{a}-1\)

\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)\)

\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-3}\right)\)\(=\dfrac{\left(a-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)

Thiếu đề nha bạn

Giải:

Đặt \(y=b\sqrt{1-x}\)

Ta có: \(\sqrt{a+y}=1+\sqrt{a-y}\)

\(\Leftrightarrow\sqrt{a+y}-\sqrt{a-y}=1\)

\(\Leftrightarrow\left(\sqrt{a+y}-\sqrt{a-y}\right)^2=1\)

\(\Leftrightarrow a+y-2\cdot\sqrt{a+y}\cdot\sqrt{a-y}+a-y=1\)

\(\Leftrightarrow2a-2\sqrt{a^2-y^2}=1\)

\(\Leftrightarrow2\sqrt{a^2-y^2}=2a-1\)

\(\Leftrightarrow\sqrt{a^2-y^2}=\dfrac{2\left(a-\dfrac{1}{2}\right)}{2}=a-\dfrac{1}{2}\)

\(\Leftrightarrow a^2-y^2=\left(a-\dfrac{1}{2}\right)^2=a^2-a+\dfrac{1}{4}\)

\(\Leftrightarrow y^2=a-\dfrac{1}{4}\)

\(\Leftrightarrow y=\sqrt{a-\dfrac{1}{4}}\)

\(\Leftrightarrow b\sqrt{1-x}=\sqrt{a-\dfrac{1}{4}}\)

\(\Leftrightarrow\sqrt{1-x}=\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\)

\(\Leftrightarrow1-x=\left(\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\right)^2\)

\(\Leftrightarrow x=1-\left(\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\right)^2\)

Vậy....................

2. 

a,  Với m\(=1\Rightarrow x^2-x=0\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b. Ta có \(\Delta=b^2-4ac=\left(-m\right)^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2\ge0\)

\(\Rightarrow\)phương trình luôn có 2 nghiệm \(x_1,x_2\)

c, Theo hệ thức Viet ta có \(\hept{\begin{cases}x_1+x_2=m\\x_1.x_2=m-1\end{cases}}\)

A=\(\frac{2.x_1x_2+3}{x_1^2+x_2^2+2\left(1+x_1x_2\right)}=\frac{2.x_1x_2+3}{\left(x_1+x_2\right)^2-2x_1x_2+2+2x_1x_2}\)

\(=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\frac{2m+1}{m^2+2}=\frac{\left(m^2+2\right)-\left(m^2-2m+1\right)}{m^2+2}\)

\(=1+\frac{-\left(m-1\right)^2}{m^2+2}\)

Ta thấy \(\frac{-\left(m-1\right)^2}{m^2+2}\le0\Rightarrow1+\frac{-\left(m-1\right)^2}{m^2+2}\le1\)

\(\Rightarrow MaxA=1\)

Dấu bằng xảy ra\(\Leftrightarrow\) \(m-1=0\Leftrightarrow m=1\)

pt\(\Leftrightarrow a-2\sqrt{a}+1+b-1-4\sqrt{b-1}+4+c-2-6\sqrt{c-2}+9=0\)(dk\(a\ge0,b\ge1,c\ge2\)

\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-1}-2\right)^2+\left(\sqrt{c-2}-3\right)^2=0\)

\(\Leftrightarrow\begin{cases}\sqrt{a}-1=0\\\sqrt{b-1}-2=0\\\sqrt{c-2}-3=0\end{cases}\)\(\Leftrightarrow\begin{cases}\sqrt{a}=1\\\sqrt{b-1}=2\\\sqrt{c-2}=3\end{cases}\)

tự giải tiếp bạn nhé