tìm x biết 7x^2-7x=0
Tìm x,biết:
a) x^2 - 4x -5 = 0
b) 4x^2 + 7x - 11 = 0
c) -7x^2 + 6x + 1 = 0
d) - 10x^2 +7x+3 = 0
a) x2 - 4x - 5 = 0
=> x2 - 5x + x - 5 = 0
=> x(x - 5) + (x - 5) = 0
=> (x + 1)(x - 5) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
b) 4x2 + 7x - 11 = 0
=> 4x2 + 11x - 4x - 11 = 0
=> x(4x + 11) - (4x + 11) = 0
=> (x - 1)(4x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)
c) -7x2 + 6x + 1 = 0
=> -7x2 + 7x - x + 1 = 0
=> -7x(x - 1) - (x - 1) = 0
=> (-7x - 1)(x - 1) = 0
=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)
d) -10x2 + 7x + 3 = 0
=> -10x2 + 10x - 3x + 3 = 0
=> -10x(x - 1) - 3(x - 1) = 0
=> (-10x - 3)(x - 1) = 0
=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)
\(a,x^2-4x-5=0\)
\(\Rightarrow x^2-5x+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
\(b,4x^2+7x-11=0\)
\(\Rightarrow4x^2-4x+11x-11=0\)
\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)
\(c,-7x^2+6x+1=0\)
\(\Rightarrow-7x^2+7x-x+1=0\)
\(\Rightarrow-7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-7x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
\(d,-10x^2+7x+3=0\)
\(\Rightarrow-10x^2+10x-3x+3=0\)
\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)
Tìm x biết
x+2\(\sqrt{7x^2}\)+7x3=0
Tìm x biết, -7x^2-x+196=0
Bài này em nên đi theo hướng giải theo delta
TÌM x biết
2x^4+7x^3+x^2-7x-3=0
MONG CÁC BẠN GIÚP MÌNH
Tìm x, biết
2(x+7) - x2 - 7x = 0
\(PT\Leftrightarrow2\left(x+7\right)-x\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-7;2\right\}\)
Tìm X biết.
a) 7x - 10 = 5x - 6
b) 3x( x - 2 ) + x - 2 = 0
c) 2x2 + 7x - 4 = 0
a) \(7x-10=5x-6\)
\(7x-5x=-6+10\)
\(2x=4\)
\(x=2\)
b) \(3x\left(x-2\right)+x-2=0\)
\(\left(x-2\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
c) \(2x^2+7x-4=0\)
\(2x^2-x+8x-4=0\)
\(x\left(2x-1\right)+2\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
7x-10=5x-6<=>7x-5x=-6+10<=>2x=4=>x=2
3x(x-2)+x-2=0<=>(x-2)(3x+1)=0<=>x-2=0=>x=2 HAY 3x+1=0=>x=-1/3
2x2+7x-4=0.
Câu cuối xem có lộn đề không nha bạn ơi!!!
a) 7x - 10 = 5x - 6
<=> 7x - 5x = -6 + 10
<=> 2x = 4
<=> x = 2
b) 3x( x - 2 ) + x - 2 = 0
<=> 3x( x - 2 ) + 1( x - 2 ) = 0
<=> ( x - 2 )( 3x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
c) 2x2 + 7x - 4 = 0
<=> 2x2 + 8x - x - 4 = 0
<=> 2x( x + 4 ) - 1( x + 4 ) = 0
<=> ( x + 4 )( 2x - 1 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{1}{2}\end{cases}}\)
tìm x biết
a,2x^2-6x+4=0
b,5x^2-10x+4=0
c,x^2+7x+12=0
d,13x^2+15x-10=0
g,7x^2-4x-1=0
tìm x
a)(3x-1)^2+2(3x-1)(2x+1)+(2x+1)^2=0
b)(7x+2)^2+(7x-2)^2-2(7x+2)(7x-2)=0
I don't now
sorry
...................
nha
a) \(\left(3x-1\right)^2+2\left(3x-1\right)\left(2x+1\right)+\left(2x+1\right)^2=0\)
\(\Leftrightarrow\)\(\left[\left(3x-1\right)+\left(2x-1\right)\right]^2=0\)
\(\Leftrightarrow\)\(\left(5x-2\right)^2=0\)
\(\Leftrightarrow\)\(5x-2=0\)
\(\Leftrightarrow\)\(x=\frac{2}{5}\)
Vậy...
b) \(\left(7x+2\right)^2+\left(7x-2\right)^2-2\left(7x+2\right)\left(7x-2\right)=0\)
\(\Leftrightarrow\)\(\left[\left(7x+2\right)-\left(7x-2\right)\right]^2=0\)
\(\Leftrightarrow\)\(4^2=0\) vô lí
Vậy pt vô nghiệm
tìm x biết
x^2+7x+8=0