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\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
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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
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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
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\(x^2+2x\sqrt{x+\frac{1}{x^2}}=8x-1\)
\(\Leftrightarrow x^2+2x\left(x+\frac{1}{x^2}\right)^2=8x-1\)
\(\Leftrightarrow x^2+2x\left(x+\frac{1}{x^2}\right)^2=7x\)
\(\Rightarrow x^2+2x\left(x+\frac{1}{x^2}\right)^2>7x\Rightarrow\)Phương trình vô nghiệm
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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
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ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !
câu 1 ) thì đúng
câu 2 sai đề
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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 \(x=1\)
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\(\frac{\sqrt{x}}{1+\sqrt{1-x}}=x^2-2x+2\Leftrightarrow\frac{\sqrt{x}-1}{1+\sqrt{1-x}}+\frac{1}{1+\sqrt{1-x}}-1=x^2-2x+1\)
\(\Leftrightarrow\frac{x-1}{\left(1+\sqrt{1-x}\right)\left(\sqrt{x}+1\right)}+\frac{-\sqrt{1-x}}{1+\sqrt{1-x}}=\left(1-x\right)^2\)
\(\Leftrightarrow\sqrt{1-x}\left[\left(\sqrt{1-x}\right)^3+\frac{\sqrt{1-x}}{\left(1+\sqrt{1-x}\right)\left(\sqrt{x}+1\right)}+\frac{1}{1+\sqrt{1-x}}\right]=0\)
\(\Leftrightarrow\sqrt{1-x}=0\Leftrightarrow x=1.\)