bài 1 tìm x:a)(x+2)^3-(x+5)(x^2+x-3)
b)(x+1)^2=2(x+1)(x+2)+(x+2)^2=9
c)(x+2)^3-(x+5)(x^2+x-3)=5
d)(12x-5)(4x-1)+(3x-7)(1-16x)=81
1) (x+5)(x+2)-3(4x-3)=(5-x)2
2) (x+2)3-(x-2)3=12x(x-1)-8
3) 3x(12x-4)-9x(4x-3)=30
4) (12x-5)(4x-1)+(3x-7)(1-16x)=81
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
Bài 1. Tìm \(x\).
a) -5(\(x\)2-3\(x\)+1)+\(x \)(1+5\(x\))=\(x-2\)
b) \(12x\)2\(-4x\)\((3x+5)\)\(=10x-17\)
c) \(-4x(x-5)+7x(x-4)-3x\)2\(=12\)
Bài 2. Tính ( Rút gọn).
a) \((x+5).(x-7)-7x.(x-3)\)
b) \(x.(x\)2\(-x-2)-(x-5).(x+1)\)
c) \((x-5).(x-7)-.(x+4).(x-3)\)
d) \((x-1).(x-2)-(x+5).(x+2)\)
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
2x mũ 2 + 3(x - 1 )( x + 1 ) = 5x ( x + 1)
x( 2x - 1 ) ( x + 5) - ( 2x mũ 2 + 1) ( x + 9/2 ) = 7/2
( 12x - 5 ) ( 4x - 1 ) + ( 3x - 7 ) ( 1 - 16x) = 81
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
=> \(2x^2+3\left(x^2-1\right)=5x^2+5x\)
=> \(2x^2+3x^2-3-5x^2-5x=0\)
=> \(-3-5x=0\)
=> \(5x=-3\Rightarrow x=-\frac{3}{5}\)
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
=> \(x\left[2x\left(x+5\right)-1\left(x+5\right)\right]-2x^2\left(x+\frac{9}{2}\right)-1\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
=> \(x\left(2x^2+10x-x-5\right)-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
=> \(2x^3+10x^2-x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
=> \(\left(2x^3-2x^3\right)+\left(10x^2-x^2-9x^2\right)+\left(-5x-x\right)-\frac{9}{2}=\frac{7}{2}\)
=> \(-6x-\frac{9}{2}=\frac{7}{2}\)
=> \(-6x=8\Rightarrow x=-\frac{8}{6}=-\frac{4}{3}\)
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
=> 12x(4x - 1) - 5(4x - 1) + 3x(1 - 16x) - 7(1 - 16x) = 81
=> 48x2 - 12x - 20x + 5 + 3x - 48x2 - 7 + 112x = 81
=> -12x - 20x + 3x + 112x + 5 - 7 = 81
=> 83x + 5 - 7 = 81
=> 83x = 81 + 7 - 5
=> 83x = 83
=> x = 1
1) \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow5x=-3\)
\(\Rightarrow x=-\frac{3}{5}\)
2) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
\(\Leftrightarrow-6x=8\)
\(\Rightarrow x=-\frac{4}{3}\)
3) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-32x+5-48x^2+115x-7=81\)
\(\Leftrightarrow83x=83\)
\(\Rightarrow x=1\)
a) 2x2 + 3( x - 1 )( x + 1 ) = 5x( x + 1 )
<=> 2x2 + 3( x2 - 1 ) = 5x2 + 5x
<=> 2x2 + 3x2 - 5x2 - 5x = 3
<=> -5x = 3
<=> x = -3/5
b) x( 2x - 1 )( x + 5 ) - ( 2x2 + 1 )( x + 9/2 ) = 7/2
<=> x( 2x2 + 9x - 5 ) - ( 2x3 + 9x2 + x + 9/2 ) = 7/2
<=> 2x3 + 9x2 - 5x - 2x3 - 9x2 - x - 9/2 = 7/2
<=> -6x - 9/2 = 7/2
<=> -6x = 8
<=> x = -8/6 = -4/3
c) ( 12x - 5 )( 4x - 1 ) + ( 3x - 7 )( 1 - 16x ) = 81
<=> 48x2 - 32x + 5 + 115x - 48x - 7 = 81
<=> 83x - 2 = 81
<=> 83x = 83
<=> x = 1
bài1:nhân
a, (x-2).(6x2-5x-3)
b,(x+2).(4x2-2x3-8x+x4+16)
c,(x2+1+x).(x2+1-x)
d,(x-y).(x4+x3y+x2y2+xy3+y4)
bài 2:rút gọn biểu thức
a,(x2-5).(x+3)+(x+4).(x-x2)
b, (12x-5).(4x-1)+(3x-7).(1-16x)-81
c, (6x-1).(3x-1)-(2x-3).(9x-1)+2
d,(3x-1).(2x+7)-(x+1).(6x-5)
Bài 1 : Tìm x biết :
a) 5|x-1| + |x-2| = |12-3x|
b) 12x+7 - 3|x-3| = 5|x+1|
Bài 2 : Tìm x biết :
a) |x-3| < 2
b) 2|x-1| - 5(2-x) > 2
c) 4|x-1| - 7|2-4x| < 5
d) 3(2-3x) + |x-3| - 2|1-x| > 3
giải chi tiết ra giúp mk nhé, cảm ơn nhiều
Tìm x
(15x-5) (4x-1) + (3x-7) (1-16x) =81
(2x+4) (x-4) +(x-5) (x-2) =3x+5 (x-4)
(8x-3) (3x+2) - (4x+7) (x+4) = (x+1) (5x-1)
Tìm x, biết:
a) (2x+2)(x-1)-(x+2)(2x+1)=0;
b)(3x+1)(2x-3)-6x(x+2)=16;
c)(12x-5)(4x-1)+(3x-7)(1-16x)=81
mn ơi giúp mik vs ạ :<
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
a) (12x-5)(4x-1)+(3x-7)(1-16x)=81
b) (2x-3)(2x+3)-(4x+1).x=1
c) \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
<=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
<=> \(83x-2=81\)
<=> \(83x=83\)
<=> \(x=1\)
b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)
<=> \(4x^2-9-4x^2-x=1\)
<=> \(-\left(9+x\right)=1\)
<=> \(9+x=-1\)
<=> \(x=-10\)
c/ \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)
<=> \(3x^2-3x^2+x-6x+2=-7\)
<=> \(-5x+2=-7\)
<=> \(-5x=-9\)
<=> \(x=\frac{9}{5}\)
Tìm x:
a)(x-1)^2+x(5-x)=8
b)x^3-3x^2+x-3
c)(12x^4-6x):6x+2x(2+x)(2-x)=7
a, <=> x2 -2x +1 + 5x -x2 =8
<=> 3x +1 =8
<=> 3x = 7
<=> x= 7/3
b, thiếu đề
c, <=> 2x3 -1 + 2x(4 -x2) = 7
<=> 2x3 + 8x -23 = 8
<=> 8x =8
<=> x=1