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Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
a) VT bạn bình phương rồi B.C.S sẽ được VT<=2
VP=3x^2-12x+12+2=3(x-2)^2+1>=2
Dấu = xảy ra khi x=2
\(\text{Đk: }1,5\le x\le2,5\)
Áp dụng bđt cauchy ta có:
\(\text{VT }\Leftrightarrow\frac{2x-3+1+1-2x+1}{2}=2\)
Mà: \(\text{VP}=3\left(x-2\right)^2+2\ge2\)
\(\text{ĐT}\Leftrightarrow x=2\)
\(\Rightarrow x=2\)
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
b,ĐK:\(-3\le x\le\frac{3}{2}\)
\(PT\Leftrightarrow x-1+4\left(\sqrt{x+3}-2\right)+2\left(\sqrt{3-2x}-1\right)=0\)
\(\Leftrightarrow x-1+\frac{4\left(x-1\right)}{\sqrt{x+3}+2}+\frac{2\left(2-2x\right)}{\sqrt{3-2x}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{4}{\sqrt{x+3}+2}-\frac{4}{\sqrt{3-2x}+1}\right)=0\)
Với \(x\ge-3\) \(\Rightarrow\frac{4}{\sqrt{x+3}+2}>0\) và \(3-2x\le9\Rightarrow-\frac{4}{\sqrt{3-2x}+1}\ge-1\)
\(\Rightarrow1+\frac{4}{\sqrt{x+3}+2}-\frac{4}{\sqrt{3-2x}+1}>0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)(tm)
c,Đk: \(x\ge2,y\ge3,z\ge5\)
pt <=> \(x-2\sqrt{x-2}+y-4\sqrt{y-3}+z-6\sqrt{z-5}+4=0\)
<=> \(\left(x-2\right)-2\sqrt{x-2}+1+\left(y-3\right)-4\sqrt{y-3}+4+\left(z-5\right)-6\sqrt{z-5}+9=0\)
<=>\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=\)0
=>\(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)(t/m)
d, \(2x+2y+2z=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}\left(đk:x,y,z\ge\frac{1}{4}\right)\)
<=> \(4x+4y+4z=2\sqrt{4x-1}+2\sqrt{4y-1}+2\sqrt{4z-1}\)
<=> \(\left(4x-1\right)-2\sqrt{4x-1}+1+\left(4y-1\right)-2\sqrt{4y-1}+1+\left(4z-1\right)-2\sqrt{4z-1}+1=0\)
<=>\(\left(\sqrt{4x-1}-1\right)^2+\left(\sqrt{4y-1}-1\right)^2+\left(\sqrt{4z-1}-1\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{4x-1}-1=0\\\sqrt{4y-1}-1=0\\\sqrt{4z-1}-1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=\frac{1}{2}\end{matrix}\right.\)(tm)