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a) \(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)
\(\Rightarrow15x-9x^2-25+15x+9\left(x^2+2x+1\right)-30=0\)
\(\Rightarrow30x-9x^2-25+9x^2+18x+9-30=0\)
\(\Rightarrow48x-46=0\)
\(\Rightarrow x=\frac{23}{24}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Rightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow8x+17=16\)
\(\Rightarrow8x=-1\)
\(\Rightarrow x=\frac{-1}{8}\)
c) \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)
\(\Rightarrow\left(y-2\right)^3-\left(y^3-3^3\right)+6\left(y^2+2y+1\right)=49\)
\(\Rightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)
\(\Rightarrow\left(y^3-y^3\right)+\left(-6y^2+6y^2\right)+\left(12y+12y\right)+\left(-8+27+6\right)=49\)
\(\Rightarrow24y+25=49\)
\(\Rightarrow24y=24\)
\(\Rightarrow y=1\)
d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)
\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)
\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)
\(\Rightarrow3y^2+12y+13=28\)
\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)
\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)
\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
\(a,\Leftrightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\\ \Leftrightarrow24y=24\Leftrightarrow y=1\\ b,\Leftrightarrow y^3+9y^2+27y+27-y^3-3y^2-3y-1=56\\ \Leftrightarrow6y^2+24y-30=0\\ \Leftrightarrow y^2+4y-5=0\\ \Leftrightarrow\left(y-1\right)\left(y+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\)
a) \(\Leftrightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)
\(\Leftrightarrow24y=24\Leftrightarrow y=1\)
b) \(\Leftrightarrow y^3+9y^2+27y+27-y^3-3y^2-3y-1=56\)
\(\Leftrightarrow6y^2+24y-30=0\)
\(\Leftrightarrow6\left(y-1\right)\left(y+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\)
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
\(y\left(2y-3\right)\left(2y-1\right)\left(y+1\right)=24\)
\(\Leftrightarrow\left[y\left(2y-1\right)\right]\left[\left(2y-3\right)\left(y+1\right)\right]=24\)
\(\Leftrightarrow\left(2y^2-y\right)\left(2y^2-y-3\right)=24\)
\(\Leftrightarrow t\left(t-3\right)=24\) (với \(t=2y^2-y\)), suy ra \(t\ge-\dfrac{1}{8}\)
\(\Leftrightarrow t^2-3t-24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{3+\sqrt{105}}{2}\left(nhận\right)\\t=\dfrac{3-\sqrt{105}}{2}\left(loại\right)\end{matrix}\right.\)
Suy ra \(2y^2-y=\dfrac{3+\sqrt{105}}{2}\)
Tới đây thì mình nghĩ bạn tìm đc y rồi đó.
\(1,\\ b,\Leftrightarrow\left(x^2+4x+4\right)+\left(y-1\right)^2=25\\ \Leftrightarrow\left(x+2\right)^2+\left(y-1\right)^2=25\)
Vậy pt vô nghiệm do 25 ko phải tổng 2 số chính phương
\(2,\\ a,\Leftrightarrow x^2-\left(y^2-6y+9\right)=47\\ \Leftrightarrow x^2-\left(y-3\right)^2=47\)
Mà 47 ko phải hiệu 2 số chính phương nên pt vô nghiệm
\(b,\Leftrightarrow\left(x-2\right)^2+\left(3y-1\right)^2=16\)
Mà 16 ko phải tổng 2 số chính phương nên pt vô nghiệm
2b,
Vì 16 ko đồng dư với 1 (mod 4) nên 16 ko phải là tổng 2 scp
Định lý Fermat về tổng của hai số chính phương – Wikipedia tiếng Việt
vô đây đọc nhé
to ko biet dau nhe
\(\left(y+3\right)^3-\left(y+1\right)^3=56\)
\(\Rightarrow[\left(y+3\right)-\left(y+1\right)][\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2-56=0\)
\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)-56=0\)
\(\Rightarrow y^2+6y+9+y^2+4y+3+y^2+2y+1-28=0\)
\(\Rightarrow3y^2+12y-15=0\)
\(\Rightarrow y^2+4y-5=0\)
\(\Rightarrow\left(y-1\right)\left(y+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y=1\\x=-5\end{cases}}\)