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1) 4x\(^2\).(5x3+2x-1)
= 20x\(^5\)+8x\(^3\)-4x\(^2\).
2) 4x\(^3\): x2
= 4x
3) ( 15x2y3-10x3y3+6xy): 5xy
= 3xy2-2x2y2+\(\dfrac{6}{5}\)
4) (5x3+14x2+12x+8 ): (x+2)
= 5x2+4x+4
5)\(\dfrac{7}{2x}\)+\(\dfrac{11}{3y^2}\)
=\(\dfrac{7.3y^2+11.2x}{6xy^2}\) =\(\dfrac{21y^2+22x}{6xy^2}\) = \(\dfrac{21+22}{6}\) =\(\dfrac{43}{6}\)
6) \(\dfrac{x}{x+2}\) +\(\dfrac{3}{\left(x+2\right)\left(4x-7\right)}\)
7)\(\dfrac{3}{x-y}\)-\(\dfrac{2x^2}{x+y}\)
= \(\dfrac{3\left(x+y\right)-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{3x+3y-2x-2y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{1}{x-y}\).
8)\(\dfrac{1}{2}\)x2y2.(2x+y)(2x-y)
= \(\dfrac{1}{2}\)x2y2.(4x2-2xy+2xy-y2)
= \(\dfrac{1}{2}\)x2y2.(4x2-y2)
= 2x4y2-\(\dfrac{1}{2}\)x2y4
9) (x-\(\dfrac{1}{2}\)).(x+\(\dfrac{1}{2}\)).(4x-1)
= x2.(4x-1)
= 4x3-x2
10)\(\dfrac{3x}{2x+6}\)+\(\dfrac{6-x}{2x^2+6x}\)
= \(\dfrac{3x}{2\left(x+3\right)}\)+\(\dfrac{6-x}{2x\left(x+3\right)}\)= \(\dfrac{3x^2+6-x}{2x\left(x+3\right)}\)=\(\dfrac{3-x}{3}\)= -x
11) x2-\(\dfrac{1}{2x-2}\)+3x+\(\dfrac{3}{1-x^2}\)
12)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)
= \(\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)= \(\dfrac{x}{x+y}\)
a)
\(A=x^2y-y+xy^2-x\)
\(A=\left(x^2y-x\right)-\left(y-xy^2\right)\)
\(A=x.\left(xy-1\right)-y.\left(1-xy\right)\)
\(A=x.\left(xy-1\right)+y.\left(xy-1\right)\)
\(A=\left(xy-1\right).\left(x+y\right)\)
Thay \(x=-5\) và \(y=2\) vào biểu thức A, ta được:
\(A=\left[\left(-5\right).2-1\right].\left[\left(-5\right)+2\right]\)
\(A=\left(-11\right).\left(-3\right)\)
\(A=33.\)
Vậy giá trị của biểu thức A tại \(x=-5\) và \(y=2\) là \(33.\)
Chúc bạn học tốt!
theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
\(A=\left(x+1\right)^2+\left(x+2\right)^2=\left(x+1\right)^2+\left(-2-x\right)^2\ge\frac{1}{2}\left(x+1-2-x\right)^2=\frac{1}{2}.1^2=\frac{1}{2}\Rightarrow A_{min}=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)
\(B=-2x^2-4\le0-4=-4\Rightarrow B_{max}=-4\Leftrightarrow x=0\)
\(C=-5x^2+10x-7=-5x^2+10x-5-2=-5\left(x-1\right)^2-2\le0-2=-2\Rightarrow C_{min}=-2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
b) \(=y^3-1+\frac{2}{3}x^3y-2xy+\frac{1}{3}x^2y^3-y^3\)
\(=\frac{2}{3}x^3y+\frac{1}{3}x^2y^3-2xy-1\)
1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)
4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
5, \(2x^3+3x^2+2x+3\)
\(=x^2\left(2x+3\right)+2x+3\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
6, \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xy^2\)
\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)
\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)
\(=xz\left(x+y\right)\left(x-z\right)\)
8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)
9, \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
10, \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
Chỗ còn lại mai làm nốt nha.
Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha
11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)
\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
12, \(x^3-7x-6\)
\(=x^3-3x^2+3x^2-9x+2x-6\)
\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
13, \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
14, \(a^4+64\)
\(=a^4+16a^2+64-16a^2\)
\(=\left(a^2+8\right)^2-16a^2\)
\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
15, \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
16, \(x^5+x-1\)
\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)
17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)
19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)
Đặt \(x^2+8x+7=a\) ta có:
(*) \(\Leftrightarrow a\left(a+8\right)+15\)
\(\Leftrightarrow a^2+8a+15\)
\(\Leftrightarrow a^2+3a+5a+15\)
\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)
\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)
Đặt \(x^2+3x+1=a\) ta có:
(*) \(\Leftrightarrow a\left(a+1\right)-6\)
\(\Leftrightarrow a^2+a-6\)
\(\Leftrightarrow a^2+3a-2a-6\)
\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)