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

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

ĐK:\(x\ge-\frac{1}{2}\)

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

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

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

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

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

Suy ra x=1

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

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

\(\Leftrightarrow\frac{4x^2-8x+3}{-x^2+2x-1}+\frac{3x+1-\frac{5}{2}}{\sqrt{3x+1}+\sqrt{\frac{5}{2}}}=\frac{-\left(4x^2-1\right)}{x^2}+\frac{x+2-\frac{5}{2}}{\sqrt{x+2}+\sqrt{\frac{5}{2}}}\)

\(\Leftrightarrow\frac{2\left(x-\frac{1}{2}\right)\left(2x-3\right)}{-x^2+2x-1}+\frac{6\left(x-\frac{1}{2}\right)}{\sqrt{3x+1}+\sqrt{\frac{5}{2}}}+\frac{2\left(x-\frac{1}{2}\right)\left(2x+1\right)}{x^2}-\frac{x-\frac{1}{2}}{\sqrt{x+2}+\sqrt{\frac{5}{2}}}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(\frac{2\left(2x-3\right)}{-x^2+2x-1}+\frac{6}{\sqrt{3x+1}+\sqrt{\frac{5}{2}}}+\frac{2\left(2x+1\right)}{x^2}-\frac{1}{\sqrt{x+2}+\sqrt{\frac{5}{2}}}\right)=0\)

Suy ra x=1/2

96 đặt\(\sqrt{x+7}+\sqrt{6-x}=a\)

=>\(a^2-13=2\sqrt{-x^2-x+42}\)

xong cậu thay vào pt là đc

1,

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

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

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

=>x=1 vì cái còn lại >0

đề sai rùi đe dung như này vì mk đã làm rồi

\(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{2x+1}}\)\(+\frac{1}{\sqrt{1-2x}}=\frac{4\sqrt{10}}{5}\)

dk \(-\frac{1}{2}< x< \frac{1}{2}\)

ap dung bdt \(\frac{1}{a}+\frac{1}{b}>=\frac{4}{a+b}\)

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

tiep tuc ap dung bdt \(a+b< =2\sqrt{a^2+b^2}\) 

\(\frac{1}{\sqrt{2x+1}}+\frac{1}{\sqrt{1-2x}}>=\frac{4}{\sqrt{2x+1}+\sqrt{1-2x}}>=\frac{4}{\sqrt{2\left(2x+1+1-2x\right)}}=2\)

lai co \(\frac{-1}{2}< x< \frac{1}{2}\Rightarrow\frac{1}{\sqrt{x+1}}>\frac{1}{\sqrt{\frac{1}{2}+1}}=\frac{\sqrt{6}}{3}\)

suy ra \(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{2x+1}}+\frac{1}{\sqrt{1-2x}}>2+\frac{\sqrt{6}}{3}>\frac{4\sqrt{10}}{5}\)

pt vo no

\(x^2+2x\sqrt{x+\frac{1}{x^2}}=8x-1\)

\(\Leftrightarrow x^2+2x\left(x+\frac{1}{x^2}\right)^2=8x-1\)

\(\Leftrightarrow x^2+2x\left(x+\frac{1}{x^2}\right)^2=7x\)

\(\Rightarrow x^2+2x\left(x+\frac{1}{x^2}\right)^2>7x\Rightarrow\)Phương trình vô nghiệm

hk như lm rồi đấy

1/ \(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)

\(\Leftrightarrow\frac{3-x}{\sqrt{5-x}}+\frac{3+x}{\sqrt{5+x}}=\frac{4}{3}\)

Đặt \(\hept{\begin{cases}\sqrt{5-x}=a\\\sqrt{5+x}=b\end{cases}}\) thì ta có:

\(\hept{\begin{cases}\frac{a^2-2}{a}+\frac{b^2-2}{b}=\frac{4}{3}\\a^2+b^2=10\end{cases}}\)

Tới đây thì đơn giản rồi nhé

2/ \(\sqrt[3]{x+\frac{1}{2}}=16x^3-1\)

\(\Leftrightarrow x+\frac{1}{2}=\left(16x^3-1\right)^3\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(8x^2+4x+1\right)\left(512x^6+64x^4-64x^3+8x^2-4x+3\right)=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

câu 2 đề sai

ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !

câu 1 ) thì đúng

câu 2 sai đề

bài 1 chắc bạn sai đề. Mình lười lắm nên link đây nhé https://diendantoanhoc.net/topic/96618-sqrtx8frac3x27x84x2/

Đặt \(a=\sqrt{2-x^2};b=\sqrt{2-\frac{1}{x^2}};c=x+\frac{1}{x}\)

xet x<0 vt < 2 căn 2<3, vt >4=>loại=>x>0=>c>=2;

ta có a+b=4-c;

a^2+b^2=4-x^2-1/x^2=6-c^2;

\(=>\hept{\begin{cases}2a+2b=8-2c\left(2\right)\\a^2+b^2=6-c^2\left(1\right)\end{cases}}\)

trừ 1 cho 2=>a^2-2a+b^2-2b=-c^2-2-2c=>a^2-2b+1+b^2-2b+1=-c^2+2c-1+1

=>\(\left(a-1\right)^2+\left(b-1\right)^2=-\left(c-1\right)^2+1\)

\(< =>\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=1\)

ta lại có (a-1)^2>=0;(b-1)^2>=0;(c-1)^2>=(2-1)^2=1=>Vế trái>=1=Vế phải, dấu bằng xảy ra<=>

\(\hept{\begin{cases}a=1\\b=1\\c=2\end{cases}< =>x=1}\)

Bạn tham khảo nhé:Điều kiện bạn tự tìm nhé

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\Leftrightarrow x=1\left(N\right)\\\frac{1}{\sqrt{2-x^2}-x+2}+\frac{1}{x\sqrt{2x^2-1}-x+2x^2}=0\left(1\right)\end{cases}}\)

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

Nhân 2 vào ta có:

\(\Leftrightarrow2x\sqrt{2x^2-1}-4x+4x^2+4+2\sqrt{2-x^2}=0\)

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

Vậy phương trình có 1 nghiệm duy nhất là \(x=1\)

Bổ sung cách độc lạ hơn nè mình vừa nghĩ ra:

Chuyển vế:

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

Ap dụng BĐT a+b<=\(\sqrt{2\left(a^2+b^2\right)}\)

Dấu = khi a=b

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

=2+2=4=VP. Dấu = xảy ra khi \(\hept{\begin{cases}\sqrt{2-x^2}=x\\\sqrt{2-\frac{1}{x^2}}=\frac{1}{x}\end{cases}< =>x=1}\)