tìm x 4.(x+1).(-x+2)+(2x-1).(2x+3)=-11 (2x+4).(3x+1).(x-2)-(-3x2 +1).(-2x+2/3)=-26/3
Những câu hỏi liên quan
Bài 1: Thực hiện phép tính:
a) 2x.(3x + 3) b) 5x.(3x2-2x + 1) c) 3x2(2x +4)
d) 5x2.(3x2 + 4x – 1) e) (x-1).(2x +3) f) (x+2).(3x-5)
Bài 2: Tìm x, biết:
a) 3x(x+1) – 3x2 = 6
b) 3x(2x+1) – (3x +1).(2x-3) = 10
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
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tìm x (2x+4).(3x+1).(x-2)-(-3x2 +1).(-2x+2/3)=-26/3
x=\(\dfrac{11}{2}\)=5\(\dfrac{1}{2}\)=5.5x=0
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Tìm x, biết:a) 2(5x-8)-3(4x-5) 4(3x-4) + 11;b)
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Tìm x, biết:
a) 2(5x-8)-3(4x-5) = 4(3x-4) + 11;
b) 2 x ( 6 x - 2 x 2 ) + 3 x 2 ( x - 4 ) = 8;
c) 2 ( x 3 - 1 ) - 2 x 2 ( x + 2 x 4 ) + ( 4 x 5 + 4 ) x = 6;
d)(2x)2(4x-2)-(x3 -8x2) = 15.
a) x = 2 7 b) x = 2.
c) x = 2 d) x = 1.
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Bài 1: Rút gọn biểu thức sau:a. 3x2(2x3- x+5) - 6x5-3x3+10x2b. -2x(x3-3x2-xx+11)-2x4+3x3+2x2-22x2xBài 2: Chứng minh biểu thức sau không phụ thuộc vào x:a. x(2x+1)-x2(x+2)+(x2-x+3)b. 4(x-6)-x2(2+3x)+x(5x-4)+3x2(x-1)Bài 3: Cho đa thức: f(x)3x2-x+1 g(x)x-1a. Tính f(x).g(x)b. Tìm x để f(x).g(x)+x2[1-3g(x)]Bài 4: Tìm x:a. dfrac{1}{4}x2-(dfrac{1}{2}x-4)dfrac{1}{2}x-14b. 2x(x-4)+3(x-4)+x(x-2)-5(x-2)3x(x-4)-5(x-4)Các bạn giúp mik giải bt nha. Cảm ơn mn nhiêu ạ.
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Bài 1: Rút gọn biểu thức sau:
a. 3x2(2x3- x+5) - 6x5-3x3+10x2
b. -2x(x3-3x2-xx+11)-2x4+3x3+2x2-22x2x
Bài 2: Chứng minh biểu thức sau không phụ thuộc vào x:
a. x(2x+1)-x2(x+2)+(x2-x+3)
b. 4(x-6)-x2(2+3x)+x(5x-4)+3x2(x-1)
Bài 3: Cho đa thức: f(x)=3x2-x+1
g(x)=x-1
a. Tính f(x).g(x)
b. Tìm x để f(x).g(x)+x2[1-3g(x)]=
Bài 4: Tìm x:
a. \(\dfrac{1}{4}\)x2-(\(\dfrac{1}{2}\)x-4)\(\dfrac{1}{2}\)x=-14
b. 2x(x-4)+3(x-4)+x(x-2)-5(x-2)=3x
(x-4)-5(x-4)
Các bạn giúp mik giải bt nha. Cảm ơn mn nhiêu ạ.
`@` `\text {Ans}`
`\downarrow`
Gửi c!
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Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
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4:
a: =>1/4x^2-1/4x^2+2x=-14
=>2x=-14
=>x=-7
b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20
=>3x^2-12x-2=3x^2-17x+20
=>5x=22
=>x=22/5
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Giải phương trình1) 2x ( x – 3 ) + 5 ( x – 3 ) 02) ( x2 – 4 ) – ( x – 2 ) ( 3 – 2x ) 03) ( 2x – 1 )2 – ( 2x + 5 )2 114) ( 2x + 1 )2 ( 3x – 5 ) 4x2 – 15) 3x2 – 5x – 8 06) ( 2x + 1 )2 ( 3x – 5 ) 4x2 – 17) 3x2 – 5x – 8 08) left|x-5right|39) left|2x-5right|3-x10) left|2x+1right|left|x-1right|11) dfrac{5x+2}{6}-dfrac{8x-1}{3}dfrac{4x+2}{5}-512) dfrac{3x+2}{2}-dfrac{3x+1}{6}2x+dfrac{5}{3}
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Giải phương trình
1) 2x ( x – 3 ) + 5 ( x – 3 ) = 0
2) ( x2 – 4 ) – ( x – 2 ) ( 3 – 2x ) = 0
3) ( 2x – 1 )2 – ( 2x + 5 )2 = 11
4) ( 2x + 1 )2 ( 3x – 5 ) = 4x2 – 1
5) 3x2 – 5x – 8 = 0
6) ( 2x + 1 )2 ( 3x – 5 ) = 4x2 – 1
7) 3x2 – 5x – 8 = 0
8) \(\left|x-5\right|=3\)
9) \(\left|2x-5\right|=3-x\)
10) \(\left|2x+1\right|=\left|x-1\right|\)
11) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
12) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
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5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
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tìm x
1. 3x - 7 = 19
2. (2x+1) . (x-3)=0
3. 3x+2/4 +1=5x-1/3
4. x/15 + 1/2 - x/50 = 5/6
5. x4 - 2x3 + 3x2 - 2x + 1=0
\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)
\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)
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Bài 9: Tìm x biết:
d) (2x - 1) . (3x + 1) + (3x + 4) . (3 - 2x) = 5
e) 0,6x (x - 0,5) - 0,3x (2x + 1,3) = 0,38
f) (c - 3x2)(x + 6) + x(3x2 + 17x) = 24
d: =>6x^2+2x-3x-1+9x-6x^2+12-8x=5
=>13=5(loại)
e: =>0,6x^2-0,3x-0,6x^2-0,39x=0,38
=>-0,69x=0,38
=>x=-38/69
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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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8 tháng 9 2021 lúc 14:22
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
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a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
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b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
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c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
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d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
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e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
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f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
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