a: ĐKXĐ: x>=-2

\(\sqrt{5x+10}=8-x\)

=>\(\begin{cases}8-x\ge0\\ \left(8-x\right)^2=5x+10\end{cases}\Rightarrow\begin{cases}x\le8\\ x^2-16x+64=5x+10\end{cases}\)

=>\(\begin{cases}-2\le x\le8\\ x^2-21x+54=0\end{cases}\Rightarrow\begin{cases}-2\le x\le8\\ \left(x-3\right)\left(x-18\right)=0\end{cases}\)

=>x=3

b: ĐKXĐ: \(4x^2+x-12\ge0\)

=>\(x^2+\frac14x-3\ge0\)

=>\(x^2+2\cdot x\cdot\frac18+\frac{1}{64}-\frac{193}{64}\ge0\)

=>\(\left(x+\frac18\right)^2\ge\frac{193}{64}\)

=>\(\left[\begin{array}{l}x+\frac18\ge\frac{\sqrt{193}}{8}\\ x+\frac18\le-\frac{\sqrt{193}}{8}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{193}-1}{8}\\ x\le\frac{-\sqrt{193}-1}{8}\end{array}\right.\)

\(\sqrt{4x^2+x-12}=3x-5\)

=>\(\begin{cases}3x-5\ge0\\ \left(3x-5\right)^2=4x^2+x-12\end{cases}\Rightarrow\begin{cases}3x\ge5\\ 9x^2-30x+25-4x^2-x+12=0\end{cases}\)

=>\(\begin{cases}x\ge\frac53\\ 5x^2-31x+37=0\end{cases}\)

\(\Delta=\left(-31\right)^2-4\cdot5\cdot37=221\) >0

=>Phương trình có hai nghiệm phân biệt là

\(\left[\begin{array}{l}x=\frac{31-\sqrt{221}}{2\cdot5}=\frac{31-\sqrt{221}}{10}\left(loại\right)\\ x=\frac{31+\sqrt{221}}{10}\left(nhận\right)\end{array}\right.\)

@Akai Haruma , @phynit giải dùm em vs ạ

a.

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1-x^2}=t\Rightarrow0\le t\le1\)

\(x^2=1-t^2\Rightarrow x^4=t^4-2t^2+1\)

Pt trở thành:

\(729\left(t^4-2t^2+1\right)+8t=36\)

\(\Leftrightarrow729t^4-1458t^2+8t+693=0\)

\(\Leftrightarrow\left(9t^2+2t-9\right)\left(81t^2-18t-77\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9t^2+2t-9=0\\81t^2-18t-77=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{82}-1}{9}\\t=\dfrac{1+\sqrt{78}}{9}\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt{1-t^2}=...\)

b.

ĐKXĐ: ...

\(-3\left(10+4x-x^2\right)-5\sqrt{10+4x-x^2}+42=0\)

Đặt \(\sqrt{10+4x-x^2}=t\ge0\)

\(\Rightarrow-3t^2-5t+42=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-\dfrac{14}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{10+4x-x^2}=3\)

\(\Leftrightarrow x^2-4x-1=0\)

\(\Leftrightarrow x=...\)

\(1+\sqrt{x^2-4x+3}-x=0\)

\(ĐK:\left\{{}\begin{matrix}\sqrt{x^2-4x+3\ge0}\\x-1\ge0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x\ge3\end{matrix}\right.\)

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

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

\(\Leftrightarrow2-2x=0\Rightarrow x=1\left(tm\right)\)

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

=>\(\sqrt{\left(x+3\right)\left(x-7\right)}=x+4\)

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

=>\(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+2\sqrt{x-4}\)

=>\(\sqrt{2x-3}=\sqrt{x-1}\)

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\) (1)

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

=>\(2\sqrt{x-1}=\frac{x+3}{2}\)

=>\(4\sqrt{x-1}=x+3\)

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

=>\(\sqrt{x-1}<1\)

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

ai giải hộ với nhanh cái mk sắp đi học òi

thui chữa òi ko cần làm đâu

a/ \(\Leftrightarrow x^2+5x-2-2\sqrt[3]{x^2+5x-2}+4=0\)

Đặt \(\sqrt[3]{x^2+5x-2}=a\)

\(a^3-2a+4=0\)

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

\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\Rightarrow x^2+5x+6=0\Rightarrow...\)

b/ ĐKXĐ:...

\(\Leftrightarrow-3\left(-x^2+4x+10\right)-5\sqrt{-x^2+4x+10}+42=0\)

Đặt \(\sqrt{-x^2+4x+10}=a\ge0\)

\(-3a^2-5a+42=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{14}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+4x+10}=3\Rightarrow x^2-4x-1=0\Rightarrow...\)

c/ ĐKXĐ: ...

\(\Leftrightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\)

Đặt \(\sqrt{x^2+3x}=a\ge0\)

\(a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+3x}=2\Rightarrow x^2+3x-4=0\)

d/ ĐKXĐ: \(-1\le x\le2\)

\(\Leftrightarrow\sqrt{3-x+x^2}=1+\sqrt{2+x-x^2}\)

\(\Leftrightarrow3-x+x^2=3+x-x^2+2\sqrt{2+x-x^2}\)

\(\Leftrightarrow2+x-x^2+\sqrt{2+x-x^2}-2=0\)

Đặt \(\sqrt{2+x-x^2}=a\ge0\)

\(a^2+a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2+x-x^2}=1\Leftrightarrow x^2-x-1=0\)

e/ \(\Leftrightarrow\sqrt{x^2-3x+3}-1+\sqrt{x^2-3x+6}-2=0\)

\(\Leftrightarrow\frac{x^2-3x+2}{\sqrt{x^2-3x+3}+1}+\frac{x^2-3x+2}{\sqrt{x^2-3x+6}+2}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{\sqrt{x^2-3x+3}+1}+\frac{1}{\sqrt{x^2-3x+6}+2}\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

ĐKXĐ: \(x\ge1\)

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

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

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

\(\Leftrightarrow x-2=0\)