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\(\left(5-xy\right)^2=25-10xy+x^2y^2\)
\(\left(3-2y\right)^2=9-12y+4y^2\)
\(\left(3+x^2\right)\left(3-x^2\right)=9-x^4\)
\(\left(5x-2y\right)\left(25x+10xy+4y^2\right)=\left(5x-2y\right)\left(5x+2y\right)=25x^2-4y^2\)\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)=\left(3x+y\right)\left(3x-y\right)=9x^2-y^2\)
a) Đặt \(f_{\left(x\right)}=0\)
\(\Leftrightarrow x^3+3x^2-2x-2=0\)
\(\Leftrightarrow x^3-x^2+4x^2-4x+2x-2=0\)
\(\Leftrightarrow x^2\left(x-1\right)+4x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+4x+4-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+2\right)^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+2=\sqrt{2}\\x+2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{2}-2\\x=-\sqrt{2}-2\end{matrix}\right.\)
Vậy: \(S=\left\{1;\sqrt{2}-2;-\sqrt{2}-2\right\}\)
b) Đặt \(G_{\left(x\right)}=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=\frac{-1}{3}\)
Vậy: \(S=\left\{-\frac{1}{3}\right\}\)
c) Đặt \(A_{\left(x\right)}=0\)
\(\Leftrightarrow2x^2-4=0\)
\(\Leftrightarrow2x^2=4\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
Vậy: \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)
d) Đặt \(h_{\left(x\right)}=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2+5x-2x-5=0\)
\(\Leftrightarrow x\left(2x+5\right)-\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{-5}{2};1\right\}\)
e) Đặt P=0
\(\Leftrightarrow3x^2+4x^2+6x+3=0\)
\(\Leftrightarrow7x^2+6x+3=0\)
\(\Leftrightarrow7\left(x^2+\frac{6}{7}x+\frac{3}{7}\right)=0\)
mà 7>0
nên \(x^2+\frac{6}{7}x+\frac{3}{7}=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{6}{14}+\frac{9}{49}+\frac{12}{49}=0\)
\(\Leftrightarrow\left(x+\frac{3}{7}\right)^2=-\frac{12}{49}\)(vô lý)
Vậy: S=∅
a,\(\left(3x^2.y^2\right).\left(-2xy^2\right)\)
\(=\left(-6\right).x^3.y^4\)
Hok tốt
\(a,\left(1+3x\right)^2=1+6x+9x^2\)
\(b,\left(2a^2+b^2\right)^2=4a^4+4a^2b^2+b^4\)
\(c,\left(\frac{x}{2}-2y\right)^2=\frac{x^2}{4}-2xy+4y^2\)
\(d,\left(x^2-0,1\right)^2=x^4-0,2x+0,01\)
\(e,\left(2x-3\right)\left(2x+3\right)=4x^2-9\)
\(g,\left(a^2+5\right)\left(5-a^2\right)=25-a^4\)
\(h,39.41=\left(40-1\right)\left(40+1\right)=40^2-1=1599\)
a) \(\left(1+3x\right)^2=1+6x+9x^2\)
b) \(\left(2a^2+b^2\right)^2=4a^4+4a^2b^2+b^4\)
c) \(\left(\frac{x}{2}-2y\right)^2=\frac{x^2}{4}-2xy+4y^2\)
d) \(\left(x^2-0,1\right)^2=x^4-0,2x^2+0,01\)
e) \(\left(2x-3\right)\left(2x+3\right)=\left(2x\right)^2-3^2=4x^2-9\)
g) \(\left(a^2+5\right)\left(5-a^2\right)=25-a^4\)