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6x^2-(2x+5)(3x-2)=7
(3x-5)*(7-5x)-(5x+2)(2-3x)=4
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11 tháng 8 2021
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
YN
31 tháng 12 2021
Answer:
\(6x^2-\left(2x+3\right)\left(3x-2\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+9x-4x-6\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+5x-6\right)=7\)
\(\Rightarrow6x^2-6x^2-5x+6=7\)
\(\Rightarrow-5x+6=7\)
\(\Rightarrow-5x=1\)
\(\Rightarrow x=\frac{-1}{5}\)
\(5x\left(12+7\right)-3x\left(80x-5\right)=-100\)
\(\Rightarrow5x.19-240x^2+15x=-100\)
\(\Rightarrow95x-240x^2+15x=-100\)
\(\Rightarrow-240x^2+110x+100=0\)
\(\Rightarrow-24x^2-11x-10=0\)
\(\Rightarrow24\left(x^2-\frac{11}{24}x+\frac{121}{2304}\right)-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2=\frac{1081}{2304}\)
\(\Rightarrow\left(x-\frac{11}{48}\right)^2=\left(\frac{\pm\sqrt{1081}}{48}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{11}{48}=\frac{\sqrt{1081}}{48}\\x-\frac{11}{48}=\frac{-\sqrt{1081}}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{1081}+11}{48}\\x=\frac{11-\sqrt{1081}}{48}\end{cases}}\)
\(\left(3x-5\right)\left(7-5x\right)-\left(5x-2\right)\left(2-3x\right)=4\)
\(\Rightarrow\left(21x-15x^2-35+25x\right)-\left(10x-15x^2-4+6x\right)-4=0\)
\(\Rightarrow36x-15x^2-35-16x+15x^2+4-4=0\)
\(\Rightarrow\left(-15x^2+15x^2\right)+\left(36x-16x\right)+\left(-35+4-4\right)=0\)
\(\Rightarrow30x-35=0\)
\(\Rightarrow x=\frac{7}{6}\)
20 tháng 7 2022
a: \(\Leftrightarrow6x^2-6x^2+4x-9x+6=7\)
=>-5x=1
hay x=-1/5
b: \(\Leftrightarrow5x\left(12x+7\right)-3x\left(80x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-240x^2+15x=-100\)
\(\Leftrightarrow-180x^2+50x+100=0\)
hay \(x\in\left\{\dfrac{5+\sqrt{745}}{36};\dfrac{5-\sqrt{745}}{36}\right\}\)
c: \(\Leftrightarrow21x-15x^2-35+25x-\left(10x-15x^2-4+6x\right)=4\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-16x+4=4\)
=>30x-31=4
=>30x=35
hay x=7/6
23 tháng 4 2023
a: =3x^3-15x^2+21x
b: =-x^3+6x^2+5x-4x^2-24x-20
=-x^3+2x^2-19x-20
c: =9x^2+15x-3x-5-7x^2-14
=2x^2+12x-19
d: =10x^2-4x+2/3
OP
29 tháng 1 2022
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
JS
15 tháng 2 2020
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
25 tháng 5 2022
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
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
6x^2-(2x+5)(3x-2)=7
6x^2 - (6x^2 + 15x -4x -10) =7
-11x = -3
x= 3/11
(3x-5)*(7-5x)-(5x+2)(2-3x)=4
21x - 35 - 15x^2 + 25x - (10x +4 - 15x^2 - 6x) =4
46x - 10x +6x = 4 + 35 +4
42x = 43
x= 42/42