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1.
ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$
PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)
\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)
\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)
2.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$
$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$
$\Leftrightarrow 2\sqrt{x}=0$
$\Leftrightarrow x=0$
Thử lại thấy thỏa mãn
Vậy $x=0$
![](https://rs.olm.vn/images/avt/0.png?1311)
4x+1−|8−x|=x−9
⇔ −|8−x|=x−9-4x-1
⇔ −|8−x|=-3x−10
⇔ |8−x|=3x+10
TH1: 8−x≥0 ⇔ x≤8
8−x=3x+10
⇔ -x-3x=10-8
⇔ -4x=2
⇔ x=\(\dfrac{-1}{2}\) (TMĐK)
TH2: 8−x<0 ⇔ x>8
x-8=3x+10
⇔ x-3x=10+8
⇔ -2x=18
⇔ x=-9 (KTMĐK)
Vậy x=\(\dfrac{-1}{2}\)
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ĐK: \(x\ge\frac{2017}{2018}\)
\(pt\Leftrightarrow2017\sqrt{2017x-2016}-2017+\sqrt{2018x-2017}-1=0\)
\(\Leftrightarrow2017\frac{2017\left(x-1\right)}{\sqrt{2017x-2016}+1}+\frac{2018\left(x-1\right)}{\sqrt{2018x-2017}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}\right)=0\)
Dễ thấy với \(x\ge\frac{2017}{2018}\Rightarrow\)\(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}>0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
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\(\frac{x+3}{2015}+\frac{x+2}{2016}+\frac{x+1}{2017}\le-3\)
\(\Leftrightarrow\frac{x+3}{2015}+1+\frac{x+2}{2016}+1+\frac{x+1}{2017}+1\le0\)
\(\Leftrightarrow\frac{x+2018}{2015}+\frac{x+2018}{2016}+\frac{x+2018}{2017}\le0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)\le0\)
Mà \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}>0\)
⇒ x + 2018 < 0 ⇔ x < - 2018
\(\frac{x+3}{2015}+\frac{x+2}{2016}+\frac{x+1}{2017}\le-3\) \(\Leftrightarrow\frac{x+2018}{2015}+\frac{x+2018}{2016}+\frac{x+2018}{2017}\le0\) \(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)\le0\)
\(\Leftrightarrow x+2018;\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2017}\) khác dấu \(\Leftrightarrow x+2018\le0\Leftrightarrow x\le-2018\)
Vậy .............
sai bạn sửa nhé :))
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Lời giải:
Trong TH này ta thêm điều kiện $x$ là số nguyên dương.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}=\frac{x}{x+1}\)
Vậy \(\frac{x}{x+1}=\frac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)
\(\Rightarrow x\sqrt{2016-x}+2017x=(x+1)\sqrt{2017-x}+2016(x+1)\)
\(\Leftrightarrow x\sqrt{2016-x}=(x+1)\sqrt{2017-x}+2016-x\)
\(\Leftrightarrow x(\sqrt{2017-x}-\sqrt{2016-x})+\sqrt{2017-x}+2016-x=0\)
\(\Leftrightarrow \frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}+\sqrt{2017-x}+(2016-x)=0\)
Hiển nhiên ta thấy:
\(\frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}>0\)
\(\sqrt{2017-x}\geq 0\)
\(2016-x\geq 0\)
Do đó pt trên vô nghiệm
Tức là không tìm đc $x$ thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^4-8\left(x-1\right)^2-9=0\)
\(\left[\left(x-1\right)^2\right]^2-2.\left(x-1\right)^2.4+16-25=0\)
\(\left[\left(x-1\right)^2-4\right]^2-5^2=0\)
\(\left[\left(x-1\right)^2-4-5\right]\left[\left(x-1\right)^2-4+5\right]=0\)
\(\left[\left(x-1\right)^2-9\right]\left[\left(x-1\right)^2+1\right]=0\)
\(\left(x-4\right)\left(x+2\right)\left[\left(x-1\right)^2+1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(x>6\Rightarrow\left(x-5\right)^{2016}>1\)(VÔ lí)
Với \(x< 5\Rightarrow\left(x-6\right)^{2016}>1\left(voli\right)\)
Với \(5< x< 6\Rightarrow0< x-5< 1\Rightarrow\left(x-5\right)^{2016}< x-5\)
Và \(-1< x-6< 0\Rightarrow\left(x-6\right)^{2016}=\left(6-x\right)^{2016}< 6-x\)
\(\Rightarrow VP< x-5+6-x=1\left(voli\right)\)
Với x=5,6 là nghiệm của pt
Vậy :..