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\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)
\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
sau bạn đăng tách ra cho mn cùng giúp nhé
a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)
b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)
c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)
d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)
e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)
f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)
Bài 1:
a) Ta có: \(\left(1-2x\right)\left(1+2x\right)+\left(2x+3\right)^2=34\)
\(\Leftrightarrow1-4x^2+4x^2+12x+9-34=0\)
\(\Leftrightarrow12x-24=0\)
\(\Leftrightarrow12\left(x-2\right)=0\)
Vì 12≠0
nên x-2=0
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x-3\right)^2+\left(3-2x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left[\left(2x-3\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};2\right\}\)
Bài 2:
a) Ta có: \(\frac{2x-5y}{x-y}-\frac{3y}{y-x}\)
\(=\frac{2x-5y}{x-y}+\frac{3y}{x-y}\)
\(=\frac{2x-5y+3y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)
b) Ta có: \(\frac{x^2+3xy}{x^2-9y^2}+\frac{5x-x^2}{x^2-3xy}\)
\(=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}+\frac{x\left(5-x\right)}{x\left(x-3y\right)}\)
\(=\frac{x}{x-3y}+\frac{5-x}{x-3y}\)
\(=\frac{x+5-x}{x-3y}=\frac{5}{x-3y}\)