Tìm x biết:
a, x(x+3)-2x-6=0
b, 4x2-1+x(2x-1)=0
Làm hộ mk nhé.Cảm ơn các cậu nhìu 😊
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a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
\(3^{x+4}=9^{2x-1}\)
\(\Rightarrow3^{x+4}=3^{4x-2}\)
\(\Rightarrow x+4=4x-2\)
\(\Rightarrow3x=6\Rightarrow x=2\)
I 2x-3 I = I x+1 I
2x-3 = x+1
x+1 - 2x+3=0
x (1-2) +1+3=0
-1x +4 =0
-1x = 0-4
-1x =-4
x = -4 : -1
x =4
Trả lời:
\(\left|2x-3\right|=\left|x+1\right|\)
\(\Rightarrow2x-3=x+1\) hoặc \(2x-3=-\left(x+1\right)\)
TH1: \(2x-3=x+1\)
\(2x-x=1+3\)
\(x=4\)
TH2: \(2x-3=-\left(x+1\right)\)
\(2x-3=-x-1\)
\(2x+x=-1+3\)
\(3x=2\)
\(x=\frac{2}{3}\)
Vậy \(x=4;x=\frac{2}{3}\)
a) 5x.(x+3/4) = 0
=> x = 0
x+3/4 = 0 => x = -3/4
b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)
\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)
\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)
\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)
\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)
\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
=> x + 2017 = 0
x = -2017
a) để 2x - 3 > 0
=> 2x > 3
x > 3/2
b) 13-5x < 0
=> 5x < 13
x < 13/5
c) \(\frac{x+3}{2x-1}>0\)
=> x + 3 > 0
x > -3
d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)
Để x+7/x+3 < 1
=> 1 + 4/x+3 < 1
=> 4/x+3 < 0
=> không tìm được x thỏa mãn điều kiện
a,x2-25-(x+5) b,mình quên mất rồi.Đợi tí nhé
(x2-25)-(x+5)=0
(x2-52)+(x-5)=0
(x-5)(x+5)+(x-5)=0
(x-5)(x+5+1)=0
x-5=0 hoặc x+5+1=0
x=0+5 hoặc x=0-5-1
x=5 hoặc x=-6
Vậy x=5 và x=-6
Giải bpt
A) (x^2+1)×(4x-2)≫0(lớn hơn hoặc =0)
B) (x-2)×x^2>0
Mog mn giúp ạ
E cần gấp
Thak mn
1) Ta có: |x+3| \(\ge\)0; |2x+y-4| \(\ge\)0
\(\Rightarrow\) |x + 3| + |2x + y - 4| \(\ge\) 0
Dấu = xảy ra khi x+3=0 và 2x+y-4 = 0 \(\Rightarrow\)x=-3; y=10
1) |x + 3| + |2x + y - 4| = 0
\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\-6+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=10\end{cases}}\)
a) Ta có: \(x\left(x+3\right)-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
b) Ta có: \(4x^2-1+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x\right)^2-1^2+x\left(2x-1\right)=0\)
\(\Rightarrow\left[\left(2x\right)^2-1^2\right]+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)+x\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1+x\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{-1}{3}\right\}\)
a, x(x+3)-2x-6=0
⇔x(x+3)-2(x+3)=0
⇔(x+3)(x-2)=0
⇔x+3=0 hoặc x-2=0
⇔x= -3 hoặc x=2
b, 4x2-1+x(2x-1)=0
⇔(2x-1)(2x+1)+x(2x-1)=0
⇔(2x-1)(2x+1+x)=0
⇔(2x-1)(3x+1)=0
⇔2x-1=0 hoặc 3x+1=0
⇔x=1/2 hoặc x= -1/3