tìm x, biết
a. 4x(x - 5) - (x - 1) (4x -3) = 5
b. ( x - 5) (x - 1) = (x - 1) (x - 2)
c. (x - 5) (-x + 4) - (x -1) (x - 1) (x + 3) = -2x2
d. (x + 1) (x2 - x + 1) - x (x2 - 3) = 4
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\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
\(A=x^2+4x-21-x^2-4x+5=-16\\ B=-2\left(4x^2+20x+25\right)-\left(1-16x^2\right)\\ B=-8x^2-40x-50-1+16x^2=8x^2-40x-51\\ C=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ D=x^3+1-\left(x^3-1\right)=2\\ E=x^3-3x^2+3x-1-x^3+1-9x^2+1=-12x^2+3x+1\)
2.
a. 3x(12x - 4) - 9x(4x - 3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 36x2 - 36x2 - 12x + 27x = 30
<=> 15x = 30
<=> x = 2
b. x(5 - 2x) + 2x(x - 1) = 15
<=> 5x - 2x2 + 2x2 - 2x = 15
<=> -2x2 + 2x2 + 5x - 2x = 15
<=> 3x = 15
<=> x = 5
a) x2 ( 5x3 - x - 1212)= 5x5-x3-1212x
b) ( 3xy - x2 + y ) 2323x2y= 6969x3y2- 2323x4y+ 2323x2y2
c) x2 ( 4x3 - 5xy + 2x ) ( -1212 xy )=(4x5-5x3y+2x3).(-1212xy)
= -4848x6y +6060x4y2-2424x4y
2/ Tìm x, biết
a) 3x( 12x - 4 ) - 9x (4x - 3 ) = 30
=> 36x2-12x-36x2+27x=30
=> -12x +27x=30
=> 15x = 30
=>x =2
b ) x( 5 - 2x ) + 2x ( x - 1 )= 15
=> 5x-2x2+2x2-2x=15
=> 3x=15
=>x=5
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
`@` `\text {dnammv}`
`a,`
`4x(x^2-x-1)-(x^2-2)(x+3)`
`= 4x^3-4x^2-4x- [x^2(x+3)-2(x+3)]`
`= 4x^3-4x^2-4x- (x^3+3x^2-2x-6)`
`= 4x^3-4x^2-4x-x^3-3x^2+2x+6`
`= 3x^3 - 7x^2-2x+6`
`b,`
`(x+5)(x+7)-7x(x+3)`
`= x(x+7)+5(x+7)-7x^2-21x`
`= x^2+7+5x+35-7x^2-21x`
`= -6x^2-16x+35`
`c,`
`x(x^2-x-2)-(x+5)(x-1)`
`= x^3-x^2-2x- [x(x-1)+5(x-1)]`
`= x^3-x^2-2x- (x^2-x+5x-5)`
`= x^3-x^2-2x - x^2 + x -5x+5`
`= x^3-2x^2- 4x+5`
`d,`
`(x+5)(x+7)-(x-4)(x+3)`
`= x(x+7)+5(x+7)- [x(x+3)-4(x+3)]`
`= x^2+7x+5x+35 - (x^2+3x-4x-12)`
`= x^2+12x+35 - x^2+x+12`
`= 13x+47`
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
a) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)
\(\Leftrightarrow-13x-3=5\)
\(\Leftrightarrow x=\frac{-8}{13}\)
b) \(\left(x-5\right)\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow x^2-6x+5=x^2-3x+2\)
\(\Leftrightarrow-3x+3=0\)
\(\Leftrightarrow x=1\)
Vậy x=1