Tìm x
2(4x-3)-3(x+5)+4(x-10) =5(x+2)
Những câu hỏi liên quan
1) 5(x-3) (x-7)-(5x+1) (x-2)= -8
2) x(x+1) (x+2)-(x+4) (3x-5)= 84-5x
3) (9x2-5) (x+3)-3x2(3x+9)=(x-5) (x+4)-x(x-11)
4) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
5) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
6) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
7) (\(\dfrac{x}{2}\)+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)+3)=0
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: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
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1) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
2) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
3) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
4) (\(\dfrac{x}{2}\)x2+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)x4+3)=0
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=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
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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=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
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3)(x2+2x+4)(2-x)+x(x-3)(x+4)-x2+24=0
\(\Rightarrow\)8-x3+x(x2+4x-3x-12)-x2+24=0
\(\Rightarrow\)8-x3+x3+4x2-3x2-12x-x2+21=0
\(\Rightarrow\)-12x+29=0
\(\Rightarrow\)-12x=-29
\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)
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Bài 5: Tìm nghiệm của các đa thức sau: Dạng 1: a) 4x + 9 b) -5x + 6 c) 7 – 2x d) 2x + 5 Dạng 2: a) ( x+ 5 ) ( x – 3) b) ( 2x – 6) ( x – 3) c) ( x – 2) ( 4x + 10 ) Dạng 3: a) x2 -2x b) x2 – 3x c) 3x2 – 4x d) ( 2x- 1)2 Dạng 4: a) x2 – 1 b) x2 – 9 c)– x 2 + 25 d) x2 - 2 e) 4x2 + 5 f) –x 2 – 16 g) - 4x4 – 25 Dạng 5: a) 2x2 – 5x + 3 b) 4x2 + 6x – 1 c) 2x2 + x – 1 d) 3x2 + 2x – 1
2 tháng 11 2021 lúc 9:44
a. (2x - 5)2 + (4x - 10)(2 + x) + x2 + 4x + 4 = 0
b. ( 3 – x2 + 5x )( x2 – 5x + 3) = 9
\(a,\Rightarrow\left(2x-5\right)^2+2\left(2x-5\right)\left(x+2\right)+\left(x+2\right)^2=0\\ \Rightarrow\left(2x-5+x+2\right)^2=0\\ \Rightarrow3x-3=0\\ \Rightarrow x=1\\ b,\Rightarrow9-\left(x^2-5x\right)^2=9\\ \Rightarrow x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
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tìm x:
a)3(2x-3)+2(2-x)=-3
b)2x(x2-2)+x2(1-2x)-x2=-12
c)3x(2x+3)-(2x+5)(3x-2)=8
d)4x(x - 1) - 3(x2-5)-x2=(x-3)-(x+4)
e)2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
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1) (\(\dfrac{1}{2}\)x + 3)*(x2- 4x- 6)
2) (6x2 -9x +15)*(\(\dfrac{2}{3}\)x+1)
3) (3x2 -x+5)*(x3+5x-1)
4) (x-1)*(x+1)*(x-2)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
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: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
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Giải các phương trình tích sau: Mng giúp em với ạ.
a) (3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0
c) 2x(x – 3) + 5(x – 3) = 0 d) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
e) (x + 2)(3 – 4x) = x2 + 4x + 4 f) x(2x – 7) – 4x + 14 = 0
g) (2x – 5)2 – (x + 2)2 = 0 h) (x2 – 2x + 1) – 4 = 0
i) 3x2 + 2x – 1 = 0 k) x2 – 5x + 6 = 0
l) x2 – 3x + 2 = 0 m) 2x2 – 6x + 1 = -3
a: (3x-2)(4x+5)=0
=>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: (2,3x-6,9)(0,1x+2)=0
=>2,3x-6,9=0 hoặc 0,1x+2=0
=>x=3 hoặc x=-20
c: =>(x-3)(2x+5)=0
=>x-3=0 hoặc 2x+5=0
=>x=3 hoặc x=-5/2
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tìm x , biết
a. 4x(x-5)-(x-1)(4x-3)=5
b. (3x-4)(x-2) = 3x(x-9)-3
c.2(x+3)-x2 -3x=0
d. 8x3-50x=0
e. (4x-30)2-3x(3-4x)
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
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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
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a) A = -3x(x-5) +3( x2 -4x) -3x-10
b) B = 4x( x2 -7x +2) – 4( x3 -7x2 +2x -5)
c) C = 5x( x2 – x) – x2( 5x-5) -15
d) D = 7( x2 -5x+3)- x( 7x-35) -14
e) E = x2 - 4x - x( x-4) -15
A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10
A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10
A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10
A = 0 + (3\(x-3x\)) - 10
A = 0 - 10
A = - 10
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5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
