Khai triển biểu thức :
1, \(\left(2x-3y\right)^3\)
2, \(\left(2x+\dfrac{9}{2}\right)^3\)
3, \(\left(x+2y\right)^3+\left(x-2y\right)^3\)
4, \(\left(2x+1\right)^3-\left(x-1\right)^3-7\cdot\left(x+1\right)^3\)
Những câu hỏi liên quan
Thu gọn các đơn thức trong biểu thức đại số sau:
C = \(\dfrac{7}{9}x^3y^2.\dfrac{6}{11}axy^3+-5bx^2y^4.-\dfrac{1}{2}axz+ax.\left(x^2y\right)^3\)
D = \(\dfrac{\left(3x4y^3\right)^2.\left(\dfrac{1}{6}x^2y\right)+\left(8x^{n-9}\right).\left(-2x^{9-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)}\)
Sao câu hỏi của bn giống của mình vậy ???
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Thu gọn các đơn thức trong biểu thức đại số sau:
C = \(\dfrac{7}{9}x^3y^2.\dfrac{6}{11}axy^3+-5bx^2y^4.-\dfrac{1}{2}axz+ax.\left(x^2y\right)^3\)
D = \(\dfrac{\left(3x4y^3\right)^2.\left(\dfrac{1}{6}x^2y\right)+\left(8x^{n-9}\right).\left(-2x^{9-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)}\) ( với axyz khác 0)
CM các biểu thức sau không phụ thuộc vào biến x,y
a) left(2x-5right)timesleft(2x+5right)-left(2x-3right)^2-12x
b) left(2y-1right)^3-2yleft(2y-3right)^2-6yleft(2y-2right)
c) left(x+3right)left(x^2-3x+9right)-left(20+x^3right)
d) 3yleft(-3y-2right)^2-left(3y-1right)left(9y^2+3y+1right)-left(-6y-1right)^2
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CM các biểu thức sau không phụ thuộc vào biến x,y
a) \(\left(2x-5\right)\times\left(2x+5\right)-\left(2x-3\right)^2-12x\)
b) \(\left(2y-1\right)^3-2y\left(2y-3\right)^2-6y\left(2y-2\right)\)
c) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(20+x^3\right)\)
d) \(3y\left(-3y-2\right)^2-\left(3y-1\right)\left(9y^2+3y+1\right)-\left(-6y-1\right)^2\)
a: \(=4x^2-25-4x^2+12x-9-12x=-34\)
b: \(=8y^3-12y^2+6y-1-2y\left(4y^2-12y+9\right)-12y^2+12y\)
\(=8y^3-24y^2+18y-1-8y^3+24y^2-18y=-1\)
c: \(=x^3+27-x^3-20=7\)
d: \(=3y\left(9y^2+12y+4\right)-27y^3+1-36y^2-12y-1\)
\(=27y^3+36y^2+12y-27y^3-36y^2-12y\)
=0
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giải giúp mik bt này vs mn!
1)left{{}begin{matrix}2x^2+y^2+x3left(xy+1right)+2ydfrac{2}{3+sqrt{2x-y}}+dfrac{2}{3+sqrt{4-5x}}dfrac{9}{2x-y+9}end{matrix}right.
2)left{{}begin{matrix}left(x+3y+1right)sqrt{2xy+2y}yleft(3x+4y+3right)left(sqrt{x+3}-sqrt{2y-2}right)left(x-3+sqrt{x^2+x+2y-4}right)4end{matrix}right.
3)left{{}begin{matrix}x-dfrac{1}{x}y-dfrac{1}{y}2yx^3+1end{matrix}right.
4)left{{}begin{matrix}sqrt{2x-3}left(y^2+2011right)left(5-yright)+sqrt{y}yleft(y-x+2right)3x+3end{matrix}right.
5...
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giải giúp mik bt này vs mn!
1)\(\left\{{}\begin{matrix}2x^2+y^2+x=3\left(xy+1\right)+2y\\\dfrac{2}{3+\sqrt{2x-y}}+\dfrac{2}{3+\sqrt{4-5x}}=\dfrac{9}{2x-y+9}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\left(x+3y+1\right)\sqrt{2xy+2y}=y\left(3x+4y+3\right)\\\left(\sqrt{x+3}-\sqrt{2y-2}\right)\left(x-3+\sqrt{x^2+x+2y-4}\right)=4\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}\sqrt{2x-3}=\left(y^2+2011\right)\left(5-y\right)+\sqrt{y}\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
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Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
Thu gọn biểu thức
1,\(\left(x-3\right)\cdot\left(x+3\right)-\left(x+1\right)^2\)
2, \(\left(2x-1\right)^2-\left(x+2\right)^2-\left(2x-\dfrac{1}{2}\right)^2\)
3,\(\left(2x+1\right)^3-\left(2x-1\right)^3-24x^2\)
4, \(\left(x-2\right)^3-\left(2x+3\right)^3-7\cdot\left(1-x\right)^3\)
1)
(x-3).(x+3) - (x+1)2
= x2 - 32 - x2 - 2x - 1
= - 2x - 10
2)
(2x - 1)2 - (x +2)2 - (2x - \(\dfrac{1}{2}\))2
= 4x2 - 4x +1 - x2 - 4x - 4 - 4x2 + 2x - \(\dfrac{1}{4}\)
= - x2 - 6x - \(\dfrac{13}{4}\)
= - ( x2 + 6x + \(\dfrac{13}{4}\) )
= - (x2 + 2.3x + 9 - \(\dfrac{23}{4}\))
= - (x + 3)2 + \(\dfrac{23}{4}\)
3)
(2x + 1)3 - (2x -1)3 - 24x2
= (2x -1 + 2)3 - (2x - 1)3 - 24x2
= (2x-1)3 + 3.(2x-1)2.2 + 3.(2x-1).22 + 23 - (2x - 1)3 - 24x2
= 6.(4x2 - 4x + 1) + 24x - 12 +8 - 24x2
= 24x2 - 24x + 6 +24x - 4 - 24x2
= 2
4)
(x-2)3 - (2x + 3)3 - 7.(1 - x)3
= x3 - 3.x2.2 + 3x.22 - 23 - 8x3 + 3.4x2.3 - 3.2x.32 + 33 - 7.(13-3x + 3x2 - x3)
= x3 - 3.x2.2 + 3x.22 - 23 - 8x3 + 3.4x2.3 - 3.2x.32 + 33 - 7 + 21x - 21x2 + 7x3
= x3 - 6x2 + 12x - 8 - 8x3 + 36x2 - 54x2 + 27 - 7 + 21x - 21x2 + 7x3
= - 45x2 + 33x + 12
= - 45(x2 - \(\dfrac{33}{45}x-\dfrac{4}{15}\))
= \(-45.\left(x^2-2.\dfrac{11}{30}.x+\dfrac{121}{900}-\dfrac{361}{900}\right)\)
= \(-45.\left(x-\dfrac{11}{30}\right)^2+\dfrac{361}{20}\)
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Thu gọn biểu thức
1,\(\left(x-2\right)^3-\left(2x+3\right)^3-7\cdot\left(1-x\right)^3\)
2,\(\left(x+5\right)\cdot\left(x^2-5x+25\right)-\left(x-2\right)\cdot\left(x^2+2x+4\right)\)
3, \(\left(2x-3\right)\cdot\left(4x^2+6x+9\right)-\left(2x+1\right)^3\)
1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)
\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)
\(=-63x^2-21x-42\)
2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)
3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)
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1.rút gọn biểu thuc P=\(\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}\) với x\(\ne-3vàx\ne3\)
2.thực hiện phép tính \(\left(2x^4-3x^3-3x^2+6x-1\right):\left(x^2-2\right)\)
\(\left(15x^4y^6-12^3y^4-18x^2y^3\right):\left(-6x^2y^2\right)\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
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\(\hept{\begin{cases}\left(x-3\right)\cdot\left(2y+5\right)=\left(2x+7\right)\cdot\left(y-1\right)\\\left(4x+1\right)\cdot\left(3y-6\right)=\left(6x-1\right)\cdot\left(2y+3\right)\end{cases}}\)