Tìm x
a, (x+7)\(^2\)-x(x-3)=12
b, x\(^2\)-3x+2=0
K
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VH
1
29 tháng 5 2022
\(\dfrac{3x^6-4x^3}{x^3}-\dfrac{\left(3x+1\right)^2}{3x+1}-\dfrac{3x^7}{x^5}=0\)
\(\Leftrightarrow3x^3-4-3x-1-3x^2=0\)
\(\Leftrightarrow3x^3-3x^2-3x-5=0\)
\(\Leftrightarrow x\simeq1,9506\)
15 tháng 2 2020
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
MN
2
12 tháng 8 2016
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
NV
10 tháng 7 2016
\(\left(\frac{2}{3}x-\frac{4}{9}\right)\left[\frac{1}{2}+\left(-\frac{3}{7}:x\right)\right]=0\)
\(\Rightarrow\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}.\frac{1}{x}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}.\frac{1}{x}=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{1}{x}=\frac{7}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}}\)
Vậy x = 2/3 , x = 6/7
M
2
17 tháng 4 2022
a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x+1-3\right)+ 4\left(x^2+x+1-3\right)=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+5\right)=0\)
\(\Leftrightarrow x^2+x+4=0\) hay \(x^2+x-2=0\)
\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{15}{4}=0\) hay \(x^2-x+2x-2=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) (pt vô nghiệm) hay\(x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=1\) hay \(x=-2\)
-Vậy \(S=\left\{1;-2\right\}\)
17 tháng 4 2022
b) \(x^3+5x^2-10x-8=0\)
\(\Leftrightarrow x^3-2x^2+7x^2-14x+4x-8=0\)
\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+7x+4\right)=0\)
\(\Leftrightarrow x=2\) hay \(x^2+2.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{33}{4}=0\)
\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{33}{4}=0\)
\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{7}{2}-\dfrac{\sqrt{33}}{2}\right)=0\)
\(\Leftrightarrow x=2\) hay \(x=\dfrac{-7-\sqrt{33}}{2}\) hay \(x=\dfrac{-7+\sqrt{33}}{2}\)
-Vậy \(S=\left\{2;\dfrac{-7-\sqrt{33}}{2};\dfrac{-7+\sqrt{33}}{2}\right\}\)
29 tháng 8 2021
a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)
\(\Leftrightarrow3x-5\ge2x-24\)
hay \(x\ge-19\)
b: Ta có: \(2\left(5-2x\right)\ge3-x\)
\(\Leftrightarrow10-4x-3+x\ge0\)
\(\Leftrightarrow-3x\ge-7\)
hay \(x\le\dfrac{7}{3}\)
\(a,\Leftrightarrow x^2+14x+49-x^2+3x=12\\ \Leftrightarrow17x=-37\Leftrightarrow x=-\dfrac{37}{17}\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(x^2+2x7+49-x^2+3x=12\Leftrightarrow17x=-37\Leftrightarrow x=\dfrac{-37}{17}\)
b) \(x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=\left(0\right)\Leftrightarrow x=1,x=2\)