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ĐK: \(x\ge8\)
Đặt \(a=\sqrt[3]{x-1}\text{ (}a\ge\sqrt[3]{7}\text{)};\text{ }b=\sqrt{x-8}\text{ (}b\ge0\text{)}\Rightarrow x=b^2+8\)
\(a^3-b^2=x-1-\left(x-8\right)=7\text{ (*)}\)
\(pt\text{ thành }a^2-2a-\left(b^2+8-5\right)b-3\left(b^2+8\right)+31=0\)
\(\Leftrightarrow\left(a^2-2a\right)-\left(b^3+3b^2+3b\right)+7=0\)
\(\Leftrightarrow\left(a-1\right)^2-\left(b+1\right)^3+a^3-b^2=0\)
Đặt \(b+1=c\text{ (}c\ge1\text{)}\)
\(pt\text{ thành }a^3-c^3+\left(a-1\right)^2-\left(c-1\right)^2=0\)
\(\Leftrightarrow\left(a-c\right)\left(a^2+ac+c^2\right)+\left(a-c\right)\left(a+c-2\right)=0\)
\(\Leftrightarrow\left(a-c\right)\left[a^2+c^2+a+c+ac-2\right]=0\)
\(\Leftrightarrow a-c=0\text{ (do }a^2+c^2+a+c+ac-2>0\text{ với mọi }a\ge\sqrt[3]{7};c\ge1\text{)}\)
\(\Leftrightarrow a=c\Leftrightarrow a=b+1\)
Thay \(b=a-1\) vào \(\left(\text{*}\right)\)ta được
\(a^3-\left(a-1\right)^2=7\Leftrightarrow\left(a-2\right)\left(a^2+a+4\right)=0\)
\(\Leftrightarrow a-2=0\text{ hoặc }a^2+a+4=0\text{ (vô nghiệm)}\)
\(\Leftrightarrow a=2\)
\(\Rightarrow\sqrt[3]{x-1}=2\Leftrightarrow x=9\)
Kết luận: \(x=9\).
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
ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !
câu 1 ) thì đúng
câu 2 sai đề
