Tìm x:
3x(2x-4)+(6x-1)(x+2)=25
Những câu hỏi liên quan
Tìm x:
3x(2x-4)+(6x-1)(x+2)=25
tìm x
1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
2) (2x+1)^3-(3x+2)^2=(2x-5)(4x^2+10x+25)+6x(2x+1)-9x^2
(3x-2)(2x-4)=1-12x²
Tìm x:
a) (x-20) mũ 2 -(x+1)(x+3)=-7
b) (3x+5)(4-3x)=0
c) x mũ 3 -9x=0
d)2/3x (x mũ 2 -4)=0
e) (2x+1)-x(2x+1)=0
f)(2x-1) mũ 2 -(2x+5) (2x-5) =18
g)x mũ 2 -25 =6x-9
Bài 3. Rút gọn các đa thức sau
a/ (2x-3)(4x^2+6x+9)- (2x+1)(4x^2 - 2x +1)
b/ (x+ 2)(x^2- 2x+4) – (x^3- 2)
c/ (3x+ 5)(9x^2 - 15x +25)- 3x(3x-1)(3x+1)
d/ x^6 - (x^2 + x +1)(x^2 - 1)(x^2 - x+ 1)
a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28
b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10
c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x
d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1
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Tìm x
2x3-50x=0
2x(3x-5)-(5-3x)=0
9(3x-2)=x(2-3x)
(2x-1)2-25=0
25x2-2=0
X2-25=6x-9
(2x-1)2-(2x+5)(2x-5)=18
\(2x\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\left(2x+1\right)\left(3x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)
\(9\left(3x-2\right)-x\left(2-3x\right)=0\)
\(9\left(3x-2\right)+x\left(3x-2\right)=0\)
\(\left(9+x\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)
\(\left(2x-1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x^2-25\right)=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm5\)
Xem thêm câu trả lời
Bài 1: Tìm x : A) 25%x+4x=35-0,75x
B) 3/4.(x-2) - 1/2.(6-2x)=1/6x + 5
C) -1/2.(3x+5) - 2/3.(9-6x)= 3/5.(x-10) - 3
D) (1/4x - 1,5) + (5/6x-3) - (5/8x-0,5)= -4,5
2) (2x+1)^3-(3x+2)^2=(2x-5)(4x^2+10x+25)+6x(2x+1)-9x^2
Tìm x
( 2x + 1 )3 - ( 3x + 2 )2 = ( 2x - 5 )( 4x2 + 10x + 25 ) + 6x( 2x + 1 ) - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - ( 9x2 + 12x + 4 ) = 8x3 - 125 + 12x2 + 6x - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - 9x2 - 12x - 4 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 - 8x3 - 3x2 - 6x + 125 = 0
⇔ -12x + 122 = 0
⇔ -12x = -122
⇔ x = 61/6
10 tháng 1 2023 lúc 19:33
bài 1: Phân tích đa thức thành nhân tử
a)\(3x^3+6x^2\)
b)\(x^2-y^2-2x+2y\)
bài 2:
a) tìm x:\(\left(2x-1\right)^2-25=0\)
b) Tìm đa thức Q biết: \(Q.\left(x^2+3x+1\right)=x^3+2x^2-2x-1\)
Gisup mik vs
Cảm ơn
Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
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Tìm x
1. (3x+5)(4-3x)=0
2. 9(3x-2)=x(2-3x)
3. 25x^2 -2=0
4. x^2- 25=6x-9
5. (2x-1)^2-(2x+5)(2x-5)=18
6. x^3-8=(x-2)^3
7. x^3-4x^2+4x=0
8. x^2- 25+2(x+5)=0
9. 2(x^2+8x+16)- x^2+4=0
10. x^2(x-2)+7x=14
(3x+5)(4-3x)=0
3x+5 =0 hoặc 4-3x=0
3x=-5 hoặc 3x=-4
x=-5/3 hoặc x=-4/3
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9(3x-2)=x(2-3x)
9(3x-2)-x(3x-2)=0
(3x-2)(9-x)=0
3x-2=0 hoặc 9-x=0
3x=2 hoặc x= -9
x =2/3 hoặc x=-9
vậy x =2/3 ; x= -9
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25x^2 - 2=0
(5x)^2 -√2^2=0
(5x-√2)(5x+√2)=0
5x=√2 hoặc 5x = -√2
x=√2/5 hoặc x= -√2/5
vậy x=√2/5 ; x=-√2/5
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