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TT
11 tháng 12 2017
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
11 tháng 12 2017
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
27 tháng 7 2020
a) 3x + 2(5 - x) = -11
=> 3x + 10 - 2x = -11
=> 3x - 2x + 10 = -11
=> x = -21
b) 3x2 - 3x(-2 + x) = 36
=> 3x2 - 3x.(-2) - 3x.x = 36
=> 3x2 + 6x - 3x2 = 36
=> 6x = 36
=> x = 6
c) x(5 - 2x) + 2x(x - 1) = 15
=> x.5 + x.(-2x) + 2x.x + 2x.(-1) = 15
=> 5x - 2x2 + 2x2 - 2x = 15
=> 3x = 15
=> x = 5
27 tháng 7 2020
Trả lời:
a,\(3x+2.\left(5-x\right)=-11\)
\(\Leftrightarrow3x+10-2x=-11\)
\(\Leftrightarrow x+10=-11\)
\(\Leftrightarrow x=-21\)
Vậy \(x=-21\)
b,\(3x^2-3x.\left(-2+x\right)=36\)
\(\Leftrightarrow3x^2+6x-3x^2=36\)
\(\Leftrightarrow6x=36\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
c, \(x.\left(5-2x\right)+2x.\left(x-1\right)=15\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=15\)
\(\Leftrightarrow3x=15\)'
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
4 tháng 12 2017
a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)
CW
18 tháng 7 2016
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt
2 tháng 7 2019
\(a,2x\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Leftrightarrow2x^2-10x-2x^2-3x=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
\(b,\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-3x^2-x^2+x+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x=\frac{7}{2}\)
\(\Leftrightarrow x=\frac{7}{4}\)
HT
22 tháng 7 2017
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
HT
22 tháng 7 2017
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
7 tháng 5 2020
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
25 tháng 8 2020
a) ( x - 5 )( 2x + 3 ) + 2x( 1 - x )
= 2x2 - 7x - 15 + 2x - 2x2
= -5x - 15
= -5( x + 3 )
b) ( 3x - 5 )2 - ( x + 5 )( 5 - x ) - 5/2( -2x )2
= 9x2 - 30x + 25 + ( x + 5 )( x - 5 ) - 5/2.4x2
= 9x2 - 30x + 25 + x2 - 25 - 10x2
= -30x
c) ( 3x + 2 )( 4 - 6x + 9x2 ) - 3x( 3x - 2 )2 + 12( -2/3 - 3x2 )
= ( 3x )3 + 23 - 3x( 9x2 - 12x + 4 ) - 8 - 36x2
= 27x3 + 8 - 27x3 + 36x2 - 12x - 8 - 36x2
= -12x
đề bài
<=> 15x - 6x2 + 6x2-10x = 20
<=> 15x-10x=20
<=> 5x=20
<=> x= 4
Vậy S={4}
x = 4 (ti ck) đúng cho mình