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Bài 1:
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=-1\)
hay x=-4/3
b: =>x=4/8+3/7=1/2+3/7=7/14+6/14=13/14
Bài 3:
BCNN(16;32;5)=160
UCLN(16;32;5)=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(\frac{1}{3}+\frac{3}{35}<\frac{x}{210}<\frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{44}{105}<\frac{x}{210}<\frac{158}{105}\)
\(\Rightarrow\frac{88}{210}<\frac{x}{210}<\frac{316}{210}\)
\(\Rightarrow x\in\left\{89;90;91;92;...;310;311;312;313;314;315\right\}\)
Câu 3:
\(\frac{5}{3}\)\(+\frac{-14}{3}\)\(<\)\(x\)\(<\)\(\frac{8}{5}+\frac{18}{10}\)
\(\Rightarrow\)\(-9\)\(<\)\(x\)\(<\)\(3,4\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{-8;-7;-6;-5;...;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt:\(f\left(x\right)=\left(x^3-2x+3\right)^{100}+\left(x^2+5x+7\right)^{90}-2\)
Ta có: \(f\left(-2\right)=\left(\left(-2\right)^3-2\left(-2\right)+3\right)^{100}+\left(\left(-2\right)^2+5\left(-2\right)+7\right)^{90}-2\)
\(=\left(-1\right)^{100}+1^{90}-2=0\)
=> x=-2 là một ngiệm của đa thức f(x)
=> \(\left(x^3-2x+3\right)^{100}+\left(x^2+5x+7\right)^{90}-2\) chia hết cho x+2
![](https://rs.olm.vn/images/avt/0.png?1311)
vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow\left(3x-1\right)^2=25\)
\(\Leftrightarrow3x-1\in\left\{5;-5\right\}\)
hay \(x\in\left\{2;-\dfrac{4}{3}\right\}\)
c: \(\Leftrightarrow\left(2x-5\right)^3=-81\)
\(\Leftrightarrow2x-5=-3\sqrt[3]{3}\)
hay \(x=\dfrac{5-\sqrt[3]{3}}{2}\)
sao ko ai tra loi het z