Tìm x biết:
(4x-5).3x-8x+10=0
Tìm x, biết:
a) ( x 2 - 4x + 16)(x + 4) - x(x + l)(x + 2) + 3 x 2 = 0;
b) (8x + 2)(1 - 3x) + (6x - l)(4x -10) = -50.
a) Thực hiện rút gọn VT = -2x – 64
Giải phương trình -2x – 64 = 0 thu được x = -32.
b) Thực hiện rút gọn VT = -62 x +12
Giải phương trình -62x + 12 = -50 thu được x = 1.
tìm x biết
a, (x^2-4x+16)(x+4)-x(x+1)(x+2)+3x^2=0
b, (8x+2)(1-3x)+(6x-1)(4x-10)=-50
Trả lời:
a, ( x2 - 4x + 16 )( x + 4 ) - x ( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 4x2 - 4x2 - 16x + 16x + 64 - x ( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
Vậy x = 32 là nghiệm của pt.
b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = - 50
<=> 8x - 24x2 + 2 - 6x + 24x2 - 60x - 4x + 10 = - 50
<=> - 62x + 12 = - 50
<=> - 62x = - 62
<=> x = 1
Vậy x = 1 là nghiệm của pt.
Câu 1 Tìm x biết
a,(3x+2)(2x-5)=(2x-5)(2x+5)
b,4x^2-8x=0
\(a,\left(3x+2\right)\left(2x-5\right)=\left(2x-5\right)\left(2x+5\right)\\ \Leftrightarrow\left(3x+2\right)\left(2x-5\right)-\left(2x-5\right)\left(2x+5\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(3x+2-2x-5\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\\ b,4x^2-8x=0\\ \Leftrightarrow4x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b: =>4x(x-2)=0
hay x=0 hoặc x=2
tìm x
2(x+3)-x^2-3x=0
2x(3x-5)=10-6x
x^3-x^2=4x^2-8x+14
\(2\left(x+3\right)-x^2-3x=0\)
<=> \(2\left(x+3\right)-x\left(x+3\right)=0\)
<=> \(\left(x+3\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
\(2x\left(3x-5\right)=10-6x\)
<=> \(2x\left(3x-5\right)-\left(10-6x\right)=0\)
<=> \(2x\left(3x-5\right)-2\left(5-3x\right)=0\)
<=> \(2x\left(3x-5\right)+2\left(3x-5\right)=0\)
<=> \(2\left(3x-5\right)\left(x+1\right)=0\)
<=> \(\orbr{\begin{cases}3x-5=0\\x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\)\(2x+6-x^2-3x=0\)
\(\Leftrightarrow\)\(-x^2-x+6=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
\(2x\left(3x-5\right)=10-6x\)
\(\Leftrightarrow\)\(6x^2-10x=10-6x\)
\(\Leftrightarrow\)\(6x^2-4x-10=0\)
\(\Leftrightarrow\)\(2\left(x+1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\3x-5=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
Vậy....
Tìm x biết
a) (x+2).(x+3) - (x-2).(x+5)=10
b) (3x+2). (2x+9) - (x+2). (8x+11)=(x+1).(3-2x)
c) 3.(2x-1).(3x-1)-(2x-3).(9x-1)=0
d) (5x-8).(4x-5)-(3x-4).(2x+12)=12
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
Tìm x biết:
4x2-8x+4=2(1-x)(x+1)
4x2-25-(2x-5)(2x+7)=0
8x2+30x+7=0
x2+3x-18=0
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
a)
4x2-8x+4=2(1-x)(x+1)
4x2-8x+4-2+2x2=0
6x2-8x+2=0
2(3x2-4x+1)=0
3x2-3x-x+1=0
3x(x-1) -(x-1)=0
(3x-1)(x-1)=0
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)
d,
x2+3x-18=0
=> x2-3x+6x-18=8
=> x(x-3)+6(x-3)=0
=> (x-3)(x+6)=0
=> \(\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
Tìm x biết :
a) (3x³ + x² – 13x + 5) : (x² + 2x – 1) = 10
b) (x⁴ – 2x² – 8) : (x – 2) = 0
c) (x²-4x) : (x²-8x+16) = 0
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
hay x=-2
1. Tìm x, biết rằng:
a, ( x+3 )( x + 1) - x ( x - 5 ) = 11
b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = -50
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)
\(\Leftrightarrow9x+3=11\)
\(\Leftrightarrow9x=11-3\)
\(\Leftrightarrow9x=8\)
\(\Leftrightarrow x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)
\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-50-12\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=\dfrac{-62}{-62}\)
\(\Leftrightarrow x=1\)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(x^2+x+3x+3-x^2+5x=11\)
\(x+8x+3=11\)
\(x+8x=8\)
\(x\left(8+1\right)=8\)
\(x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(-62x+12=-50\)
\(-62x=-62\)
\(x=1\)
tìm x biết :
a.(6x5-3x2):3x-(4x2+8x):4x=5
b.x3+6x2+11x+6=0
b.\(x^3+6x^2+11x+6=0\)
\(\Leftrightarrow x^3+x^2+5x^2+5x+6x+6=0\)
\(\Leftrightarrow x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+3x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\)\(x+1=0\)hoặc \(x+2=0\)hoặc \(x+3=0\)
\(\Leftrightarrow\)...... tự viết nha bn