Khai triển tích \(\left(x+2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x^2-2x+4\right)\)
Chỉ cần kết quả thôi ko cần lời giải, cảm ơn m.n nhìu
Những câu hỏi liên quan
Baif1 : Phân tích đa thức sau thành nhân tử : \(f\left(x\right)=x^5+y^5-\left(x+y\right)^5\)
Bài 2: Giai phương trình ; \(\left(x-6\right)^4+\left(y-8\right)^4=16\)
Bài 3 : giải phương trình \(\left(x^2+3x-4\right)^3+\left(2x^2-5x+3\right)^3=\left(3x^2-2x-1\right)^3\)
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
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A,left(frac{x+1}{x-1}-frac{x-1}{x+1}right):left(frac{1}{x+1}-frac{x}{1-x}+frac{2}{x^2-1}right)frac{4x}{left(x+1right)^2}B,frac{2+x}{2-x}:frac{4x^2}{4-4x+x^2}cdotleft(frac{2}{2-x}-frac{4}{8+x^2}cdotfrac{4-2x+x^2}{2-x}right)frac{1}{2x}C,left[left(frac{3}{x-y}+frac{3x}{x^2-y^2}right):frac{2x+y}{x^2+2xy+y^2}right]cdotfrac{x-y}{3}xyChứng minh đẳng thức ( tìm x)mọi người giải dùm mình cảm ơn
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\(A,\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)=\frac{4x}{\left(x+1\right)^2}\)
\(B,\frac{2+x}{2-x}:\frac{4x^2}{4-4x+x^2}\cdot\left(\frac{2}{2-x}-\frac{4}{8+x^2}\cdot\frac{4-2x+x^2}{2-x}\right)=\frac{1}{2x}\)
\(C,\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right):\frac{2x+y}{x^2+2xy+y^2}\right]\cdot\frac{x-y}{3}=xy\)
Chứng minh đẳng thức ( tìm x)
mọi người giải dùm mình cảm ơn
a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
=
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M.n làm ơn giúp mink nha, cảm ơn!!!!
Chứng minh rằng với mọi số nguyên x,y thì:
a) \(x\left(x^2-2x\right)+\left(x-2x\right)\) chia hết cho x - 2
b) \(x^3y^2-3yx^2+xy\) chia hết cho xy
c) \(x^3y^2-3x^2y^3+xy^2\) chia hết cho \(x^2-3xy+1\)
a) \(x\left(x^2-2x\right)+\left(x-2x\right)=x^2\left(x-2\right)+x\left(x-2\right)=\left(x-2\right)\left(x^2+x\right)⋮x-2\forall x,y\in Z\)
b) \(x^3y^2-3yx^2+xy=xy\left(x^2y-3x+1\right)⋮xy\forall x,y\in Z\)
c) \(x^3y^2-3x^2y^3+xy^2=xy^2\left(x^2-3xy+1\right)⋮\left(x^2-3xy+1\right)\forall x,y\in Z\)
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BT1: Khai triển
\(d,\left(x+2\right)\left(x^2-2x+4\right)\)
\(e,\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)
\(=x^3+2^3\)
\(=x^3+8\)
e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)
\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)
\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)
\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)
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d: (x+2)(x^2-2x+4)
=(x+2)(x^2-x*2+2^2)
=x^3+8
e: (1/4-x/5)(1/16+x/20+x^2/25)
=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]
=1/64-x^3/125
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Rút gọn: \(A=\frac{\sqrt{1+\sqrt{1-x^2}}.\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{2+\sqrt{1-x^2}}\)
Giải giúp mk... Mk cần gấp.
Giải các phương trình sau:
f. 5 – (x – 6) = 4(3 – 2x)
g. 7 – (2x + 4) = – (x + 4)
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
i. \(\left(x-2^3\right)+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x=2x^3-16\)
<=>\(8x=-16\)
<=>\(x=-2\)
i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)
<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(6x^2-2x-10=0\)
<=>\(3x^2-x-5=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>\(x=\dfrac{1}{5}\)
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f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
<=>\(8x=-16\)
<=>x=-2
i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(9x+6=0\)
<=>x=\(\dfrac{-2}{3}\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>
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1) \(\left(3-x^2\right)+6-2x=0\)
2) \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
3) \(x^2-6x+4\left(x-6\right)=0\)
4) \(\left(x+1\right)\left(2x-3\right)=x\left(x+1\right)\)
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
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2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
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3) Ta có: \(x^2-6x+4\left(x-6\right)=0\)
\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy: S={6;-4}
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giải hệ phương trìnha)hept{begin{cases}left(x+5right)left(y-2right)left(x+2right)left(y-1right)left(x-4right)left(y+7right)left(x-3right)left(y+4right)end{cases}}b)hept{begin{cases}frac{1}{x+y}-frac{2}{x-y}2frac{5}{x+y}-frac{4}{x-y}3end{cases}}c)hept{begin{cases}4x^2+y^2132x^2-y^2-7end{cases}}d)hept{begin{cases}2xy+23x5y-frac{2}{x}4end{cases}}e)hept{begin{cases}2sqrt{x-1}+3sqrt{y-2}53sqrt{x-1}-sqrt{y-2}2end{cases}}MỌI NGƯỜI GIÚP MK LM MẤY BÀI NÀY NHA MK CẦN GẤP LẮM LUÔN
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giải hệ phương trình
a)\(\hept{\begin{cases}\left(x+5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{cases}}\)
b)\(\hept{\begin{cases}\frac{1}{x+y}-\frac{2}{x-y}=2\\\frac{5}{x+y}-\frac{4}{x-y}=3\end{cases}}\)
c)\(\hept{\begin{cases}4x^2+y^2=13\\2x^2-y^2=-7\end{cases}}\)
d)\(\hept{\begin{cases}2xy+2=3x\\5y-\frac{2}{x}=4\end{cases}}\)
e)\(\hept{\begin{cases}2\sqrt{x-1}+3\sqrt{y-2}=5\\3\sqrt{x-1}-\sqrt{y-2}=2\end{cases}}\)
MỌI NGƯỜI GIÚP MK LM MẤY BÀI NÀY NHA MK CẦN GẤP LẮM LUÔN
Ôi trời nhiều thía ? làm từng câu một ha !
a \(\hept{\begin{cases}\left(x+5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}xy-2x+5y-10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+3y=8\\3x-y=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y-3x+9y=16+24\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\8y=40\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7\\y=5\end{cases}}\)
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b, ĐKXĐ \(x\ne\pm y\)
Đặt \(\frac{1}{x+y}=a\) và \(\frac{1}{x-y}=b\)(a và b khác 0)
Ta có hệ \(\hept{\begin{cases}a-2b=2\\5a-4b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b-2a+4b=3-4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\3a=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=-\frac{1}{3}\\b=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+y}=-\frac{1}{3}\\\frac{1}{x-y}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y=-3\\x-y=-\frac{6}{7}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y-x+y=-3+\frac{6}{7}\\x-y=-\frac{6}{7}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2y=-\frac{15}{7}\\x-y=-\frac{6}{7}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-\frac{27}{14}\\y=-\frac{15}{14}\end{cases}}\)
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c,\(\hept{\begin{cases}4x^2+y^2=13\\2x^2-y^2=-7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x^2+y^2+2x^2-y^2=13-7\\2x^2-y^2=-7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}6x^2=6\\2x^2-y^2=-7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=9\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\pm1\\y=\pm3\end{cases}}\)
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Dùng hằng đẳng thức để triển khai và thu gọn:
a) \(x\left(x-1\right).\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
b) \(\left(x-1\right)^3-\left(x+2\right).\left(x^2-2x+4\right)+3.\left(x+4\right).\left(x-4\right)\)
c) \(3x^2.\left(x+1\right).\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right).\left(x^4+x^2+1\right)\)
a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2-x-x^2+x-1\right)\)
\(=\left(x+1\right)\cdot\left(-1\right)\)
\(=-1\left(x+1\right)\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)
\(=x^3-1-x^3-8+12x-12\)
\(=-21+12x\)
c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)
\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)
\(=0\)
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