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Từ đề bài, ta có các trường hợp sau:
TH1: Cả 3 thừa số đều dương:
Khi đó biểu thức trở thành:
\(\left(x-2\right)+\left(x-3\right)+\left(x-4\right)=2\)
\(\Rightarrow\left(x+x+x\right)-\left(2+3+4\right)=2\)
\(\Rightarrow3x-9=2\)
\(\Rightarrow3x=11\)
\(\Rightarrow x=\frac{11}{3}\)
Do \(\frac{11}{3}-4=-\frac{1}{3}< 0\) ( mâu thuẫn với điều kiện các thừa số đều dương ) nên ta loại.
![](https://rs.olm.vn/images/avt/0.png?1311)
X2(x+2)+4(x+2)=0
=>(x2+4)(x+2)=0
=>x2+4=0 hoặc x+2=0
=>x2=-4 hoặc x=-2
Mà x2 phải ra kết quả là số dương
suy ra x=-2
\(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\Rightarrow\left(x^2+4\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-4\\x=-2\end{cases}}}\)
mà \(x^2\ge0\Rightarrow x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|5x-4|=|x+2|
TH1: 5x-4=x+2⇒4x=6⇒x=3/2
TH2: 5x-4=-x-2⇒ 6x=2⇒x=1/3
Vậy ....
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\4-5x=x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x-1)(x-2)=0
⇒\(\left\{{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|x2 - |x - 1|| = x2 + 2
=> \(\orbr{\begin{cases}x^2-\left|x-1\right|=x^2+2\\x^2-\left|x+1\right|=-x^2-2\end{cases}}\)
=> \(\orbr{\begin{cases}\left|x-1\right|=-2\left(loại\right)\\\left|x+1\right|=2x^2+2\end{cases}}\)
=> \(\orbr{\begin{cases}x+1=2x^2+2\\x+1=-2x^2-2\end{cases}}\)
=> \(\orbr{\begin{cases}2x^2-x+1=0\\2x^2+x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2\left(x^2-\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{2}=0\\2\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)+\frac{5}{2}=0\end{cases}}\)(
=> \(\orbr{\begin{cases}2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}=0\\2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}=0\end{cases}}\)(loại)
=> ko có giá trị x thõa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\frac{9}{2}=5-\frac{x}{5}\)
\(x+\frac{x}{5}=5-\frac{9}{2}\)
\(x\left(1+\frac{1}{5}\right)=\frac{10}{2}-\frac{9}{2}\)
\(x\cdot\left(\frac{5}{5}+\frac{1}{5}\right)=\frac{1}{2}\)
\(x\cdot\frac{6}{5}=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{6}{5}\)
\(x=\frac{5}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)
\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)
\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|2x+3\right|=x+2\)
\(\Leftrightarrow\hept{\begin{cases}2x+3=x+2\\2x+3=-x-2\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-x=2-3\\2x+x=-3-2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\3x=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=-\frac{5}{3}\end{cases}}\)
x=2 để gttđ x-2+x=2=2-2+2=2
tick nha