4- |2x+3|=-5
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tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.
![](https://rs.olm.vn/images/avt/0.png?1311)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ 3-2x+4+6x=x+7+3x
⇔-2x+6x-x-3x=0
⇔0x=0 (Vô số nghiệm)
2/-6(1,5-2x)=3(-15+2x)
⇔-9+12x=-45+6x
⇔6x+36=0
⇔6(x+6)=0
⇔x+6=0
⇔x=-6
Vậy S ϵ {-6}
3/ 3(2x-5)+5(x-1)=4(x+1)
⇔6x-15+5x-5=4x+4
⇔7x=24
⇔x=\(\dfrac{24}{7}\)
Vậy S ϵ {\(\dfrac{24}{7}\)}
1) Ta có: \(3-2x+4+6x=x+7+3x\)
\(\Leftrightarrow4x+7=4x+7\)
\(\Leftrightarrow4x+7-4x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
2) Ta có: \(-6\cdot\left(1.5-2x\right)=3\left(-15+2x\right)\)
\(\Leftrightarrow-9+12x=-45+6x\)
\(\Leftrightarrow12x-9+45-6x=0\)
\(\Leftrightarrow6x+36=0\)
\(\Leftrightarrow6x=-36\)
hay x=-6
Vậy: S={-6}
3) Ta có: \(3\left(2x-5\right)+5\left(x-1\right)=4\left(x+1\right)\)
\(\Leftrightarrow6x-15+5x-5=4x+4\)
\(\Leftrightarrow11x-20-4x-4=0\)
\(\Leftrightarrow7x-24=0\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\dfrac{24}{7}\)
Vậy: \(S=\left\{\dfrac{24}{7}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)
\(=x^3+125-x^3+8x^2-16x+16x\)
\(=8x^2+125\)
\(\left|2x+3\right|=4-\left(-5\right)\)
\(\left|2x+3\right|=9\)
\(\orbr{\begin{cases}2x+3=9\\2x+3=-9\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=-12\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-6\end{cases}}}\)
Vậy:\(x\in\left\{3;-6\right\}\)
Chúc em học tốt^^
\(4-\left|2x+3\right|=-5\)
=>\(2x+3=4-\left(-5\right)=9\)
TH1 :
\(2x+3=9\)
\(2x\) \(=9-3=6\)
\(x\) \(=6:2=3\)
TH2
\(2x+3=-9\)
\(2x\) \(=-9-3=-12\)
\(x\) \(=-12:2=-6\)
Vậy \(x=3\) hoặc \(x=-6\)
K mk nha, mk nhanh nha
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