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NHO TICK NHA
Nhan xet x=0 la 1 nghiem cua phuoong trinh
Xet x>0=>\(\sqrt[5]{x-1}+\sqrt[3]{x+8}>-1+2=1>-x^3+1\)
Xet x<0=>\(\sqrt[5]{x-1}+\sqrt[3]{x+8}<-1+2=1<-x^3+1\)
Vay x=0
de bai sai ban oi Phai la -x^3+1 chu
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\) = 5
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)
Nếu \(\sqrt{x-1}\ge2\Rightarrow\left|\sqrt{x-1}-2\right|=\sqrt{x-1}-2\Rightarrow\sqrt{x-1}-2+\sqrt{x-1}+3=5\)
\(\Rightarrow2\sqrt{x-1}=4\Leftrightarrow x=5\)
Nếu \(0\le\sqrt{x-1}< 2\Rightarrow\left|\sqrt{x-1}-2\right|=2-\sqrt{x-1}\Rightarrow2-\sqrt{x-1}+\sqrt{x-1}+3=5\)
\(\Leftrightarrow2+3=5\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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NGUYỄN MINH TÀI Ok bí thì cx đừng gắt,t giải đoạn đó cho
\(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
\(VT=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\)
\(\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\("="\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x-1}\le3\Leftrightarrow4\le x-1\le9\)
\(\Leftrightarrow5\le x\le10\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)}^2+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Làm nốt nhé :v
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x+1=27
=>2x=26
=>x=13
b: =>\(\sqrt[3]{x+5}=x+5\)
=>x+5=(x+5)^3
=>(x+5)(x+4)(x+6)=0
=>x=-5;x=-4;x=-6
c: =>2-3x=-8
=>3x=10
=>x=10/3
d: =>\(\sqrt[3]{x-1}=x-1\)
=>(x-1)^3=(x-1)
=>x(x-1)(x-2)=0
=>x=0;x=1;x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\)
\(\Leftrightarrow11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{11x-5-2x^2}\)
\(\Leftrightarrow121\left(5-x\right)+176\sqrt{\left(5-x\right)\left(2x-1\right)}+64\left(2x-1\right)=576+144\sqrt{11x-5-2x^2}\)\(+9\left(11x-5-2x^2\right)\)
\(\Leftrightarrow605-121x+176\sqrt{11x-5-2x^2}+128x-64=576+144\sqrt{11x-5-2x^2}\)\(+99x-18x^2\)
\(\Leftrightarrow176\sqrt{11x-5-2x^2}-144\sqrt{11x-5-2x^2}=531+99x-18x^2-541-7x\)
\(\Leftrightarrow32\sqrt{11x-5-2x^2}=-10+92x-18x^2\)
\(\Leftrightarrow16\sqrt{11x-5-2x^2}=-5+46x-9x^2\)
\(\Leftrightarrow256\left(11x-5-2x^2\right)=25+2116x^2+81x^4-460x+90x^2-823x^3\)
\(\Leftrightarrow2816x-1280-512x^2=25+2206x^2+81x^4-460x-823x^3\)
\(\Leftrightarrow9\left(364x-145-302x^2-9x^4+92x^3\right)=0\)
\(\Leftrightarrow-9x^4+92x^3-302x^2+364x-145=0\)
\(\Leftrightarrow-\left(x-1\right)\left(9x^3-83x^2+219x-145\right)=0\)
\(\Leftrightarrow-\left(x-1\right)\left(x-1\right)\left(9x^2-74x+145\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\left(9x-29\right)\left(x-5\right)=0\Leftrightarrow\)x=1; x=29/9; x=5
\(\Leftrightarrow11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{11x-5-2x^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)