3. Tìm x, biết:
a) 2x + 8 ≤ 0
b) 4x-7 ≥ 2x -5
c) (2x-8)(15-3x)>0
d) (10-2x)(8+2x)≤0
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LH
1
13 tháng 12 2023
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
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
19 tháng 8 2021
a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
9 tháng 7 2018
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
9 tháng 7 2018
a)4(x+2)-7(2x-1)+9(3x-4)=30 b)2(5x-8)-3(4x-5)=4(3x-4)+11
<=>4x+8-14x+7+27x-36=30 <=>10x-16-12x+15=12x-16+11
<=>17x-21=30 <=> -14x=-4 <=>x=2/7
<=>17x=51
<=>x=3
JS
15 tháng 2 2020
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)