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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)
1) \(\left(x-5\right)\left(x+7\right)-7x\left(x+3\right)\)
\(=x^2+7x-5x-35-7x^2-21x\)
\(=-6x^2-19x-35\)
2) \(\left(x+5\right)\left(x+7\right)-\left(x-4\right)\left(x+3\right)\)
\(=x^2+5x+7x+35-\left(x^2+3x-4x-12\right)\)
\(=x^2+12x+35-x^2+x+12\)
\(=13x+47\)
3) \(\left(2x-3\right)\left(x+4\right)+\left(-x+1\right)\left(x-2\right)\)
\(=2x^2+8x-3x-12-x^2+2x+x-2\)
\(=x^2+8x-14\)
![](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`
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2^7+\left(x-3^7\right)=5^7-4^7\)
=>\(128+x-2187=78125-16384\)
=>\(x-2059=61741\)
=>\(x=61741+2059=63800\)
c: \(7^2-\left(x+15\right)=5\cdot2^2\)
=>49-(x+15)=5*4=20
=>x+15=29
=>x=14
d: 7^3-7(13-x)=14
=>343-7(13-x)=14
=>7(13-x)=343-14=329
=>13-x=47
=>x=13-47=-34
![](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)
Đề chép đúng không bạn? Nếu không nhầm thì số hạng thứ 3 phải là \(\dfrac{7}{\left(x+10\right)\left(x+17\right)}\)
Bạn chỉ việc phân tích kiểu \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{1}{x+2}-\dfrac{1}{x+5}\)
Các số hạng sau phân tích y hệt là rút gọn được hết.
Kết quả x=15
th1: x>=7 => x-7+x-5=3 <=> 2x=15 <=>15/2 (t/m đk)
th2: x<7<=> 7-x+x-5=3 <=> 2=3 => loại
=> x=15/2
=> |x - 7| + x = 3 + 5 = 8
=> |x - 7| = 8 - x
- Nếu x < 0 thì |x - 7| = -(x - 7) = -x + 7 = 8 - x
=> -x + 7 = 8 + (-x). Vì -x + 7 < 8 + (-x) nên x không tồn tại.
- Nếu x \(\ge\) 0 thì |x - 7| = x - 7 = 8 - x
=> x + (-7) = 8 + (-x)
=> x = \(\frac{15}{2}\)