a. (x-2)(2...
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a. (x-2)(2 x-1)=5(2-x)
b.
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H9
HT.Phong (9A5)
CTVHS
14 tháng 4 2023
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) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
17 tháng 9 2020
Bài 1.
a) 2x2 + 3( x - 1 )( x + 1 ) - 5x( x + 1 )
= 2x2 + 3( x2 - 1 ) - 5x2 - 5x
= 2x2 + 3x2 - 3 - 5x2 - 5x
= -5x - 3
b) 4( x - 1 )( x + 5 ) - ( x - 2 )( x + 5 ) - 3( x - 1 )( x + 2 )
= 4( x2 + 4x - 5 ) - ( x2 + 3x - 10 ) - 3( x2 + x - 2 )
= 4x2 + 16x - 20 - x2 - 3x + 10 - 3x2 - 3x + 6
= 10x - 4
Bài 2.
a) ( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) = 0
<=> -5x2 - 2x + 16 + 4( x2 - x - 2 ) + 2( x2 - 4 ) = 0
<=> -5x2 - 2x + 16 + 4x2 - 4x - 8 + 2x2 - 8 = 0
<=> x2 - 6x = 0
<=> x( x - 6 ) = 0
<=> x = 0 hoặc x = 6
b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 0
<=> x2 + 5x + 6 - ( x2 + 3x - 10 ) = 0
<=> x2 + 5x + 6 - x2 - 3x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8
Bài 3.
A = ( n2 + 3n - 1 )( n + 2 ) - n3 + 2
= n3 + 2n2 + 3n2 + 6n - n - 2 - n3 + 2
= 5n2 + 5n
= 5n( n + 1 ) chia hết cho 5 ( đpcm )
B = ( 6n + 1 )( n + 5 ) - ( 3n + 5 )( 2n - 1 )
= 6n2 + 30n + n + 5 - ( 6n2 - 3n + 10n - 5 )
= 6n2 + 31n + 5 - 6n2 - 7n + 5
= 24n + 10
= 2( 12n + 5 ) chia hết cho 2 ( đpcm )
F
17 tháng 9 2020
bài 1:a,\(2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=-3-5x\)
b.\(4\left(x-1\right)\left(x+5\right)-\left(x-2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=4\left(x^2+4x-5\right)-\left(x^2+3x-10\right)-3\left(x^2+x-2\right)\)
\(=4x^2+16x-20-x^2-3x+10-3x^2-3x+6\)
\(=10x-4\)
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2+2x-2x-4\right)=0\)
\(-2x+16-5x^2+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
ND
Nguyễn Đức Trí
VIP
25 tháng 8 2023
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
QA
19 tháng 7 2021
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
2 tháng 11 2017
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
22 tháng 1 2018
Bài 1:
a)|x+1|-5=20-(-5)
\(\Leftrightarrow\)|x+1|=30
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+1=30\\x+1=-30\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=29\\x=-31\end{matrix}\right.\)
Vậy x=29 hoặc x=-31.
b)(-7)-|x+1|=32+(-11)
\(\Leftrightarrow\)|x+1|=28
\(\Leftrightarrow\left[{}\begin{matrix}x+1=28\\x+1=-28\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=27\\x=-29\end{matrix}\right.\)
Vậy x=27 hoặc x=-29.
`Answer:`
a. `(x-2)(2x-1)=5(2-x)`
`<=>(x-2)(2x-1)-5(2-x)=0`
`<=>(x-2)(2x-1)+5(x-2)=0`
`<=>(x-2)(2x-1+5)=0`
`<=>2(x-2)(x+2)=0`
`<=>(x-2)(x+2)=0`
`<=>x-2=0` hoặc `x+2=0`
`<=>x=2` hoặc `x=-2`
b. \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\left(ĐKXĐ:x\ne\pm5\right)\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x^2+25x}{\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow x^2+10x+25-\left(x-10x+25\right)=x^2+25x\)
\(\Leftrightarrow x^2+10x+25-x+10x-25-x^2-25x=0\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(10x-x+10x\right)+\left(25-25\right)=0\)
\(\Leftrightarrow19x=0\)
\(\Leftrightarrow x=0\)