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`@` `\text {Ans}`
`\downarrow`
`a,`
`2/5 + x = 2/7`
`=> x = 2/7 -2/5`
`=> x= - 4/35`
Vậy, `x=-4/35`
`b,`
`x + 3/5 = -2/5`
`=> x = -2/5 - 3/5`
`=> x=-1`
Vậy, `x=-1`
`c, `
`x - 8/5 = 3/7`
`=> x=3/7 + 8/5`
`=> x=71/35`
Vậy, `x=71/35`
`d,`
`3/5 - x = 7/3`
`=> x=3/5 - 7/3`
`=> x=-26/15`
Vậy, `x=-26/15`
a) \(\dfrac{2}{5}+x=\dfrac{2}{7}\)
\(\Rightarrow x=\dfrac{2}{7}-\dfrac{2}{5}\)
\(\Rightarrow x=-\dfrac{4}{35}\)
b) \(x+\dfrac{3}{5}=-\dfrac{2}{5}\)
\(\Rightarrow x=-\dfrac{2}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=-1\)
c) \(x-\dfrac{8}{5}=\dfrac{3}{7}\)
\(\Rightarrow x=\dfrac{3}{7}+\dfrac{8}{5}\)
\(\Rightarrow x=\dfrac{71}{35}\)
d) \(\dfrac{3}{5}-x=\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{7}{3}\)
\(\Rightarrow x=-\dfrac{26}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-(x+3/5)+4/5
=1/7-2x-3/5+4/5
=-2x+12/35
b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)
\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)
-3/5<x<1/7
nên x-1/7<0; x+3/5>0
\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)
c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0
\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-(x+3/5)+4/5
=1/7-2x-3/5+4/5
=-2x+12/35
b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)
\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)
-3/5<x<1/7
nên x-1/7<0; x+3/5>0
\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)
c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0
\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {dnammv}`
`a,`
`4x(x^2-x-1)-(x^2-2)(x+3)`
`= 4x^3-4x^2-4x- [x^2(x+3)-2(x+3)]`
`= 4x^3-4x^2-4x- (x^3+3x^2-2x-6)`
`= 4x^3-4x^2-4x-x^3-3x^2+2x+6`
`= 3x^3 - 7x^2-2x+6`
`b,`
`(x+5)(x+7)-7x(x+3)`
`= x(x+7)+5(x+7)-7x^2-21x`
`= x^2+7+5x+35-7x^2-21x`
`= -6x^2-16x+35`
`c,`
`x(x^2-x-2)-(x+5)(x-1)`
`= x^3-x^2-2x- [x(x-1)+5(x-1)]`
`= x^3-x^2-2x- (x^2-x+5x-5)`
`= x^3-x^2-2x - x^2 + x -5x+5`
`= x^3-2x^2- 4x+5`
`d,`
`(x+5)(x+7)-(x-4)(x+3)`
`= x(x+7)+5(x+7)- [x(x+3)-4(x+3)]`
`= x^2+7x+5x+35 - (x^2+3x-4x-12)`
`= x^2+12x+35 - x^2+x+12`
`= 13x+47`