tìm x biết 0 ,25x^3+x^2+x=0
Tìm x biết:
a. x3 – 25x = 0 b. 3x(x- 2) – x + 2 = 0
c. x2 – 4x - 5 = 0 d.x3 – x2 + 3x – 3 = 0
e. x3 + 27 + ( x + 3)( x – 9) = 0
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
1, Tìm x, biết:
A) 2x(x-7)+5x-35=0
B) x(x-3)-7x+21=0
C) x^3-x^2-25x+25=0
Tìm x biết:
a) (x+2)^2 - 9 = 0
b) 25x^2 - 10x + 1 = 0
c) x^2 + 14x + 49 = 0
d) (2x-1)^2 + (x+3)^2 - 5(x+7) (x-7) = 0
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
Tìm x biết
a) 25x^2 -1-(5x-1)(x+2) = 0
b) (2x-3)-(3-2x)(x-1) = 0
c) 9 -4x^2-(6+4x)(x-5) = 0
b) ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0
<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0
<=> ( 2x - 3 )( 1 + x - 1 ) = 0
<=> x( 2x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Vậy .....
a, 25x^2 - 1 - (5x -1)(x+2)=0
=> (5x)^2 - 1 + (5x-1)(x+2) = 0
=> (5x-1)(5x+1) + (5x-1)(x+2) = 0
=> (5x-1)(5x+1+x+2) = 0
=> (5x-1)(6x+3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
a) 25x2 - 1 - ( 5x - 1 )( x + 2 ) = 0
<=> ( 5x - 1 )( 5x + 1 ) - ( 5x - 1 )( x + 2 ) = 0
<=> ( 5x - 1 )( 5x + 1 - x - 2) = 0
<=> ( 5x - 1 )( 4x - 1 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{4}\end{cases}}}\)
Vậy .......
Tìm x, biết 25 x - 2. 10 x + 4 x = 0
A. x = 1 B. x = -1
C. x = 2 D. x = 0
Tìm x, biết 25 x - 2 . 10 x + 4 x = 0
A. x = 1 B. x = -1
C. x = 2 D. x = 0
tìm x biết
c) x( 2x - 3 ) - 2( 3 - 2x) =0
d) 25x2 - 36 =0
c) x( 2x - 3 ) - 2( 3 - 2x) =0
\(\Leftrightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=\frac{3}{2}\end{array}\right.\)
d) 25x2 - 36 =0
\(\Leftrightarrow\left(5x\right)^2-6^2=0\)
\(\Leftrightarrow\left(5x-6\right)\left(5x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-6=0\\5x+6=0\end{array}\right.\)
\(\Leftrightarrow x=\pm\frac{6}{5}\)
a) \(x\left(2x-3\right)-2\left(3-2x\right)=0\)
=> \(\left(2x-3\right)\left(x+2\right)=0\)
=>\(\left[\begin{array}{nghiempt}2x-3=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-2\end{array}\right.\)
b) \(25x^2-36=0\)
\(\Leftrightarrow\left(5x-6\right)\left(5x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-6=0\\5x+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{6}{5}\\x=-\frac{6}{5}\end{array}\right.\)
Tìm x biết 4 - 25x^2 =0 Tính giá trị của biểu thức A=x^3 - 3x^2 + 3x - 1
1) \(4-25x^2=0\)
\(\Rightarrow\left(2-5x\right)\left(2+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=2\\5x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
2) Tính thì phải cho giá trị của x.
\(A=x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
\(=\left[{}\begin{matrix}\left(\dfrac{5}{2}-1\right)^3=\dfrac{27}{8}\\\left(-\dfrac{5}{2}-1\right)^3=-\dfrac{343}{8}\end{matrix}\right.\)
Tìm x,biết x3-25x=0
x3-25x=0
=> x(x2-25)=0
=> x(x2-52)=0
=> x(x-5)(x+5)=0
=> x=0 hoặc x-5=0 hoặc x+5=0
=> x=0 hoặc x=5 hoặc x=-5
x3-25x=0
<=>x.(x2-25)=0
<=>x.(x-5)(x+5)=0
<=>x=0 hoặc x-5=0 hoặc x+5=0
<=>x=0 hoặc x=5 hoặc x=-5
x3-25x=0 ( x>=0) => x*(x2 - 25) =0
Xét 2 trường hợp => x = 0;5