ĐKXĐ: \(x\ge1\).

Phương trình đã cho tương đương:

\(\sqrt{x+3}+\sqrt{x-1}=\dfrac{8}{\sqrt{4x^4-12x^3+9x^2+16}-\left(2x^2-3x\right)}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-1}=\dfrac{\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)}{2}\)

\(\Leftrightarrow\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)-2\sqrt{x+3}-2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(\sqrt{4x^4-12x^3+9x^2+16}-2\sqrt{x+3}\right)+\left(2x^2-3x-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\dfrac{4x^4-12x^3+9x^2-4x+4}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{4x^4-12x^3+9x^2-4x+4}{2x^2-3x+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3-4x^2+x-2\right)\left(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}\right)=0\).

Do \(x\ge1\) nên ta có \(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}>0\).

Do đó \(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\left(TMĐK\right)\\4x^3-4x^2+x-2=0\left(1\right)\end{matrix}\right.\).

Giải phương trình bậc 3 ở (1) ta được \(x=\dfrac{\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}{\sqrt[6]{279936}}+\dfrac{1}{\sqrt[6]{7776}\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}+\dfrac{1}{3}\approx1,157298106\left(TMĐK\right)\).

Vậy...

Vì trong bài làm của mình có một số dòng khá dài nên bạn có thể vào trang cá nhân của mình để đọc tốt hơn!

xem lại đề câu 1đi nhé 

b)\(\frac{1}{x+\sqrt{x^2+x}}+\frac{1}{x-\sqrt{x^2+x}}=x\)

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

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

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

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

Dễ thấy: x=0 ko là nghiệm nên \(x+2=0\Rightarrow x=-2\)

c)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)

\(\Leftrightarrow\frac{\left(2x+4\right)-4\left(2-x\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}\)

\(\Leftrightarrow\frac{2\left(3x-2\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}\)

\(\Leftrightarrow\frac{2\left(3x-2\right)}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\frac{2}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{4}{\sqrt{9x^2+16}}\right)=0\)

\(\Leftrightarrow x=\frac{2}{3}\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

a, ĐKXĐ: \(x\le2\)

\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)

b, ĐKXĐ: \(x\ge-1\)

\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)

c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow4-2x=25\)

hay \(x=-\dfrac{21}{2}\)

c: \(\Leftrightarrow\left|2x+3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

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

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

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

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

=>\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

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

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

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

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

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

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

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

a,ĐK: x\(\ge\)1

⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)

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

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha