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1) \(\left|x-2\right|=x\)
th1 : \(x\ge2\) thì \(\left|x-2\right|=x\Leftrightarrow x-2=x\Leftrightarrow-2=0x\)(vô lí)
th2 : \(x< 2\) thì \(\left|x-2\right|=x\Leftrightarrow2-x=x\Leftrightarrow2=2x\Leftrightarrow x=1\)(tmđk)
vậy \(x=1\)
b)
\(\left(x+\dfrac{1}{2}\right)^{20}\ge0\)
\(\left(x-\dfrac{1}{3}\right)^{40}\ge0\)
Mà:
\(\left(x+\dfrac{1}{2}\right)^{20}+\left(x-\dfrac{1}{3}\right)^{40}< 1\)
\(\Leftrightarrow x\in\left\{\varnothing\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\frac{1}{2}=\frac{x}{3}\)
\(\Rightarrow3x+\frac{3}{2}=x\)
\(\Rightarrow x-3x=\frac{3}{2}\)
\(\Rightarrow-2x=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}:-2\)
\(\Rightarrow x=\frac{3}{2}.-\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)
Vậy \(x=-\frac{3}{4}\)
\(x+\frac{1}{2}=\frac{x}{3}\)
\(x=\frac{x}{3}-\frac{1}{2}\)
\(\frac{6x}{6}=\frac{2x}{6}-\frac{3}{6}\)
\(\Rightarrow6x=2x-3\)
\(6x-2x=3\)
\(4x=3\)
\(x=\frac{3}{4}\)
\(\frac{1-2x}{9}=\frac{1}{1-2x}\)
\(\Rightarrow\left(1-2x\right).\left(1-2x\right)=1.9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
\(\frac{4}{5}=\frac{12}{x}=\frac{y}{20}=\frac{8.\left(y-x\right)}{z}\)
\(\Rightarrow\frac{4}{5}=\frac{12}{x}\)
\(4x=60\)
\(x=15\)
\(\frac{4}{5}=\frac{y}{20}\)
\(y5=80\)
\(y=16\)
\(\frac{4}{5}=\frac{8\left(y-x\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{8.\left(80-15\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{520}{z}\)
\(\Leftrightarrow4z=2600\)
\(z=650\)
Vậy ... ( bạn tự kết luận nhé )
chúc bạn học tốt !!!
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1, \(\frac{x+1}{2}=\frac{x}{3}\)
<=> 3(x+1)=2x
<=> 3x+3-2x=0
<=> x+3=0
<=> x=-3
2, \(\frac{1-2x}{9}=\frac{1}{1-2x}\left(x\ne\frac{1}{2}\right)\)
<=> (1-2x)2=9
<=> \(\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\left(tmđk\right)}\)
Vậy x={-1;2}
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(2,\frac{9}{x}=\frac{2}{5}-\frac{7}{20}\)
\(\Rightarrow\frac{9}{x}=\frac{1}{20}\)
\(\Rightarrow x=9.20\)
\(\Rightarrow x=180\)
\(\frac{x}{5}=\frac{5}{6}+\left(-\frac{19}{30}\right)\)
\(\frac{\Rightarrow x}{5}=\frac{1}{5}\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Ta có: \(\left|12-3x\right|-20=10\)
\(\Rightarrow\left|12-3x\right|=30\)
\(\Rightarrow12-3x=\pm30\)
Nếu \(12-3x=30\Rightarrow3x=-18\Rightarrow x=-6\)
Nếu \(12-3x=-30\Rightarrow3x=42\Rightarrow x=14\)
2. Ta có: \(\left(x+1\right)\left(y-2\right)=13=1.13=13.1=\left(-13\right).\left(-1\right)=\left(-1\right)\left(-13\right)\)
x+1 | 1 | 13 | -13 | -1 |
x | 0 | 12 | -14 | -2 |
y-2 | 13 | 1 | -1 | -13 |
y | 15 | 3 | 1 | -11 |
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\hept{\begin{cases}x+y=40\\x-y=20\end{cases}}\)
\(\Rightarrow x+y+x-y=60\)
\(\Rightarrow2x=60\)
\(\Rightarrow x=30\)
\(\Rightarrow y=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c/
\(\left(4y+1\right)\left(8-x\right)=20=1.20=5.4\)
-Nhận xét: 4y+1 là số chia 4 dư 1 nên ta có các trường hợp sau:
\(+4y+1=1\text{ và }8-x=20\Leftrightarrow\left(x;y\right)=\left(-12;0\right)\)
\(+4y+1=5\text{ và }8-x=4\Leftrightarrow\left(x;y\right)=\left(4;1\right)\)
Kết luận: \(\left(x;y\right)=\left(-12;0\right);\left(4;1\right)\)
Cách làm tương tự, đối với các câu a, b, ta chỉ thêm một bước là phân tích 1 vế thành nhân tử, 1 vế là số nguyên.
\(\text{a) }xy=x-y\Leftrightarrow\left(x+1\right)\left(y-1\right)=-1\)
\(\text{b) }x\left(y+2\right)+y=1\Leftrightarrow\left(x+1\right)\left(y+2\right)=3\)
minh ko biet lam bai nay nhung minh biet lam bai khac