tìm x
(2x-3)3 - (2x + 3)(4x2 - 1) = -24
Những câu hỏi liên quan
Rút gọn biểu thức:
a) (x + 2)(x – 2) – (x + 1)2
b) (2x – 1)(4x2 + 2x + 1) – (2x + 1)( 4x2 – 2x + 1)
3. Tìm x biết:
a) (x + 2)(x2 – 2x + 4) – x(x2 – 2) = 15
b) (x – 1)3 – x(x2 – 3x – 4) = 13
thanks
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
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Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
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Bài 3: Tìm x
a) (2x+3)2−4x2=10
b) (x+1)2−(2+x)(x−2)=0
c) (5x−1)(1+5x)=25x2−7x+15
d) (4−x)2−16=0
e) 3x2−12x=0
g) x2−8x−3x+24=0
e: \(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
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Tìm x :
a) (2x-1)3-4x2 . (2x-3)=5
b) (x+1)3-(x-1)3-6.(x+1)2=-10
c) (x+2)3-x2.(x+6)=4
a) \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\Leftrightarrow6x=6\Leftrightarrow x=1\)
b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)^2=-10\)
\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x^2+2x+1+x^2-2x+1+\left(x^2-1\right)\right)\right]-6\left(x^2+2x+1\right)=-10\)
\(\Leftrightarrow2\left(3x^2+1\right)-6x^2-12x-6=-10\)
\(\Leftrightarrow6x^2+2-6x^2-12x-6=-10\)
\(\Leftrightarrow-12x-4=-10\Leftrightarrow12x=-6\Leftrightarrow x=\dfrac{1}{2}\)
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Tìm x, biết:
a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1)-3x2=54
b)(x-3)3-(x-3)(x2+6x+9)+6(x+1)2+3x2=-33
\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)
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Bài 3: Rút gọn các biểu thức sau:
1) ( x+ 3)(x2 -3x + 9) - (x3 + 54)
2) (2x + y)(4x2 + 2xy + y2 ) - (2x – y)(4x2 + 2xy + y2 )
3) (x – 1)3 – (x + 2)(x2 -2x +4) +3(x +4)(x – 4)
4) x(x + 1)(x - 1) – (x + 1)(x2 – x +1)
5) 8x3 - 5 (2x + 1)(4x2 – 4x + 1)
6) 27 + (x – 3)(x2 +3x + 9)
7) (x – 1)3 – (x +2)(x2 -2x + 4) +3(x +4)(x -4)
8) (x – 2)3 +6( x – 1)2 –(x +1)(x2 -x +1) +3x
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
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\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
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tìm x biết |4x2+|3x+2||=4x2+2x+3|4x2+|3x+2||=4x2+2x+3
tập hợp các giá trị x thỏa mãn là {.....}
Tìm x biết:
a, 16x² – 9(x + 1)²= 0
b, x2 (x – 1) – 4x2 + 8x – 4 = 0
c, x(2x – 3) – 2(3 – 2x) = 0
d, (x – 3)(x² + 3x + 9) – x(x + 2)(x – 2) = 1
e, 4x² + 4x – 6 = 2
f, 2x² + 7x + 3 = 0
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
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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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Bài 1: Thực hiện phép tính: a) 2x.(3x2 – 5x + 3) b) (-2x-1).( x2 + 5x – 3 ) – (x-1)3c) (2x – y).(4x2 + 2xy + y2) d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2 e) (x3 – 3x2 + x – 3) : (x – 3)Bài 2: Tìm x, biết:a) 5x(x – 1) 10 (x – 1); b) 2(x + 5) – x2 – 5x 0; c) x3 - x 0; d) (2x – 1)2 – (4x – 3)2 0 e) (5x + 3)(x – 4) – (x – 5)x (2x – 5)(5+2x )Bài 3: Chứng minh r...
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Bài 1: Thực hiện phép tính:
a) 2x.(3x2 – 5x + 3) b) (-2x-1).( x2 + 5x – 3 ) – (x-1)3
c) (2x – y).(4x2 + 2xy + y2) d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2
e) (x3 – 3x2 + x – 3) : (x – 3)
Bài 2: Tìm x, biết:
a) 5x(x – 1) = 10 (x – 1); b) 2(x + 5) – x2 – 5x = 0;
c) x3 - x = 0; d) (2x – 1)2 – (4x – 3)2 = 0
e) (5x + 3)(x – 4) – (x – 5)x = (2x – 5)(5+2x )
Bài 3: Chứng minh rằng giá trị của biểu thức không phụ thuộc vào giá trị của biến.
a) x(3x + 12) – (7x – 20) + x2(2x – 3) – x(2x2 + 5).
b) 3(2x – 1) – 5(x – 3) + 6(3x – 4) – 19x.
Bài 4: Phân tích đa thức thành nhân tử.
a) 10x(x – y) – 8(y – x) b) (3x + 1)2 – (2x + 1)2
c) - 5x2 + 10xy – 5y2 + 20z2 d) 4x2 – 4x +4 – y2
e) 2x2 - 9xy – 5y2 f) x3 – 4x2 + 4 x – xy2
Bài 5: Tìm giá trị nhỏ nhất của biểu thức
a) A = 9x2 – 6x + 11 b) B = 4x2 – 20x + 101
Bài 6: Tìm giá trị lớn nhất của biểu thức
a) A = x – x2 b) B = – x2 + 6x – 11
a) 2x.(3x2 – 5x + 3)
=2x3-10x2+6x
b(-2x-1).( x2 + 5x – 3 ) – (x-1)3
=-2x3 - 10x2 + 6x - x2 - 5x + 3 - x3 + 3x2 - 3x + 1
= -3x3 - 8x2 - 2x + 4
d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2
=2x2-3xy+5y2
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