a,5(2x+3)+/2(2x+3)/+/2x+3/=16
b,/x^2+/6x-2/=x^2+4
Những câu hỏi liên quan
a,x^2+2x/(x+1)^2+3-x^2-2x/(x-1)^2+3=16/x^4+4x^2+16
b,x^2/x^2+2x+2+x^2/x^2-2x+2=5(x^2-5)/x^4+4+25/4
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
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làm phép chia
1) (x^6 - 2x^4 + 6x^3 - 4x^2) : 6x^2
2) (-2x^5 = 3x^2 - 4x^3) : 2x^2
3) (15x^3 - 10x^2 + x - 2) : (x - 2)
4) (2x^4 -3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)
b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)
c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)
d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)
c,\(\dfrac{x^2+2}{x^3+1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
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a) (2x +1)(3 – x)(4 - 2x) = 0 b)2x(x – 3) + 5(x – 3) = 0
c) (x2 – 4) – (x – 2)(3 – 2x) = 0 d) x2 – 5x + 6 = 0
e) (2x + 5)2 = (x + 2)2 f) 2x3 + 6x2 = x2 + 3x
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
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Bài 1: Tính:
a) x2(x-2x3); b) (x2+1)(5-x); c) (x-2)(x2+3x-4); d) (x-2)(x-x2+4); e) (x2-1)(x2+2x); f) (2x-1)(3x+2)(3-x)
Bài 2: Tính:
a) (x-2y)2; b) (2x2+3)3; c) (x-2)(x2+2x+4); d) (2x-1)3
Bài 3: Rút gọn biểu thức:
a) (6x+1)2+(6x-1)2-2(1+6x)(6x-1); b) 3(22+1)(24+1)(28+1)(216+1); c) x(2x2-3)-x2(5x+1)+x2; d) 3x(x-2)-5x(1-x)-8(x2-3)
Bài 4: Tính nhanh:
a) 1012; b) 97.103; c) 772+232+77.46; d) 1052-52; e) A (x-y)(x2+xy+y2)+2y3 tại x dfrac{2}{3} và y dfrac{1}{3}
Đọc tiếp
Bài 1: Tính:
a) x2(x-2x3); b) (x2+1)(5-x); c) (x-2)(x2+3x-4); d) (x-2)(x-x2+4); e) (x2-1)(x2+2x); f) (2x-1)(3x+2)(3-x)
Bài 2: Tính:
a) (x-2y)2; b) (2x2+3)3; c) (x-2)(x2+2x+4); d) (2x-1)3
Bài 3: Rút gọn biểu thức:
a) (6x+1)2+(6x-1)2-2(1+6x)(6x-1); b) 3(22+1)(24+1)(28+1)(216+1); c) x(2x2-3)-x2(5x+1)+x2; d) 3x(x-2)-5x(1-x)-8(x2-3)
Bài 4: Tính nhanh:
a) 1012; b) 97.103; c) 772+232+77.46; d) 1052-52; e) A= (x-y)(x2+xy+y2)+2y3 tại x= \(\dfrac{2}{3}\) và y= \(\dfrac{1}{3}\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
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BT: Viết biểu thức sau về dạng bình phương hoặc dạng tích.
a, x4 - 8x2 + 16
b, (4x+5)2 - (5x+4)2
c, (2x-3)2 - 2(2x-3).(x+2) + (-x-2)2
a, \(x^4-8x^2+16=\left(x^2-4\right)^2\)
b, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=\left(1-x\right)\left(9x+9\right)=9\left(1-x\right)\left(1+x\right)=9\left(1-x^2\right)\)
c, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)
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a) \(x^4-8x^2+16=\left(x^2-4\right)^2\)
b) \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)c) \(\left(2x-3\right)^2-2.\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)
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a) x+1/2x-6-4/2x-6 b) 3x-4/6x+3-x-5/6x+3
c) x-1/x-3-3x-8/3-x+3-2x/x-3 d) 3/x+5-5/x-7
e) 3/x+5-5/x-7 f) 2/x-2+3/x+2+5x-18/x2-4
Bài 1: Rút gọn các biểu thức sau:
a, A = (x-2).(2x-1) - 2x (x+3)
b, B = (3x-2).(2x+1) - (6x-1).(x+2)
c, C = 6x.(2x+3) - (4x-1).(3x-2)
d, D = (2x+3).(5x-2)+(x+4).(2x-1) - 6x.(2x-3)
Bài 2: Chứng tỏ rằng các đa thức không phụ thuộc vào biến.
a, 2x(3x-5).(x+11) - 3x.(2x+3).(x+7)
b, (x2+5x-6).(x-1) - (x+2).(x2-x+1) - x(3x-10)
c, (x2+x+1).(x-1) - x2(x+1) + x2 - 5
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
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Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
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Bài 2:
a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)
\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)
\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)
\(=-117x\)
b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)
\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)
\(=x^3+x^2-x+6-x^3-x^2+x-2\)
=4
c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)
\(=x^3-1-x^3-x^2+x^2-5\)
=-6
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Xem thêm câu trả lời
Tìm x , biết :
a) (3x -1)(2x+7) -(x+1)(6x-5) =16
b) (2x +3)2-2(2x+3)(2x-5)+(2x-5)2= x2+6x+64
c) (x4+2x3+10x-25): (x2+5)=3
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-6\)
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a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=6\)
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a) (3x−1)(2x+7)−(x+1)(6x−5)=16(3x−1)(2x+7)−(x+1)(6x−5)=16
⇔(6x2+21x−2x−7)−(6x2−5x+6x−5)−16=0⇔(6x2+21x−2x−7)−(6x2−5x+6x−5)−16=0
⇔6x2+21x−2x−7−6x2+5x−6x+5−16=0⇔6x2+21x−2x−7−6x2+5x−6x+5−16=0
⇔18x−18=0⇔18x−18=0
⇔18x=18⇔18x=18
⇔x=18:18⇔x=18:18
⇔x=1⇔x=1
Vậy x=1x=1
b) (2x+3)2−2(2x+3)(2x−5)+(2x−5)2=x2+6x+64(2x+3)2−2(2x+3)(2x−5)+(2x−5)2=x2+6x+64
⇔[(2x+3)−(2x−5)]2−(x2+6x+64)=0⇔[(2x+3)−(2x−5)]2−(x2+6x+64)=0
⇔(2x+3−2x+5)2−x2−6x−64=0⇔(2x+3−2x+5)2−x2−6x−64=0
⇔82−x2−6x−64=0⇔82−x2−6x−64=0
⇔64−x2−6x−64=0⇔64−x2−6x−64=0
⇔−x2−6x=0⇔−x2−6x=0
⇔x(−x−6)=0⇔x(−x−6)=0
⇔[x=0−x−6=0⇔[x=0−x−6=0
⇔[x=0−x=6⇔[x=0−x=6
⇔[x=0x=−6⇔[x=0x=−6
Vậy x=0x=0 hoặc x=−6
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