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:

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

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

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

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

\(\Leftrightarrow...\)

a: Ta có: \(\sqrt{x^2-x+3}+7=10\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)

\(\Leftrightarrow x^2-4x+8=4\)

\(\Leftrightarrow x-2=0\)

hay x=2

\(\Leftrightarrow x\cdot\dfrac{3}{7}-1-\dfrac{3}{7}x^2+x=0\)

\(\Leftrightarrow-\dfrac{3}{7}x^2+\dfrac{10}{7}x-1=0\)

\(\Leftrightarrow3x^2-10x+7=0\)

=>(x-1)(3x-7)=0

=>x=1 hoặc x=7/3

thì v nè

a)\(1+\sqrt{3x+1}=3x\)\(\Leftrightarrow\sqrt{3x+1}=3x-1\Leftrightarrow3x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow3x-1=9x^2-6x+1\Leftrightarrow9x^2-6x+1-3x+1=0\)

\(\Leftrightarrow9x^2-9x+2=0\Leftrightarrow9x^2-6x-3x+2=0\)

\(\Leftrightarrow3x\cdot\left(3x-2\right)-\left(3x-2\right)=0\Leftrightarrow\left(3x-1\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-1=0\\3x-2=0\end{cases}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=\frac{2}{3\left[\right]}\end{array}\right.}\)

b. \(\frac{\sqrt{5x+7}}{x+3}=4\)

ĐKXĐ: \(x\ge-\frac{7}{5}\)

\(\Leftrightarrow\sqrt{5x+7}=4\left(x+3\right)\\ \Leftrightarrow\left(\sqrt{5x+7}\right)^2=\left[4\left(x+3\right)\right]^2\\ \Leftrightarrow5x+7=16\left(x^2+6x+9\right)\\ \Leftrightarrow5x+7=16x^2+96x+144\\ \Leftrightarrow16x^2+96x-5x+144-7=0\\ \Leftrightarrow16x^2+91x+137=0\\ \Leftrightarrow\left(4x\right)^2+2.4x.\frac{91}{8}+\frac{8281}{64}+\frac{487}{64}=0\\ \Leftrightarrow\left(4x+\frac{91}{8}\right)^2+\frac{487}{64}=0\left(1\right)\)

Mà \(\left(4x+\frac{91}{8}\right)^2\ge0\forall x\Rightarrow\left(4x+\frac{91}{8}\right)^2+\frac{487}{64}\ge\frac{487}{64}>0\forall x\)

\(\Rightarrow\) phương trình (1) không xảy ra.

Vậy không cógiá trị nào của x thỏa mãn phương trình.

Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10