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Câu 20:
Ta có: \(\widehat{A}-\widehat{B}=40^0\Rightarrow\widehat{B}=\widehat{A}-40^0\)
\(\widehat{A}=2\widehat{C}\Rightarrow\widehat{C}=\frac{\widehat{A}}{2}\)
Vì AB//CD (gt) \(\Rightarrow\widehat{A}+\widehat{D}=180^0\)(hai góc trong cùng phía)\(\Rightarrow\widehat{D}=180^0-\widehat{A}\)
Tứ giác ABCD \(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\Rightarrow\widehat{A}+\left(\widehat{A}-40^0\right)+\frac{\widehat{A}}{2}+\left(180^0-\widehat{A}\right)=360^0\)
Và đến đây bạn dễ dàng tìm được góc A và từ đó suy ra được góc D.
Câu 29: Ta có:
\(\hept{\begin{cases}xy+x+y=3\\yz+y+z=8\\xz+x+z=15\end{cases}}\Leftrightarrow\hept{\begin{cases}xy+x+y+1=4\\yz+y+z+1=9\\xz+x+z+1=16\end{cases}\Leftrightarrow}\hept{\begin{cases}x\left(y+1\right)+\left(y+1\right)=4\\y\left(z+1\right)+\left(z+1\right)=9\\x\left(z+1\right)+\left(z+1\right)=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=4\\\left(y+1\right)\left(z+1\right)=9\\\left(z+1\right)\left(x+1\right)=16\end{cases}}\)
Đặt \(\hept{\begin{cases}x+1=a\\y+1=b\\z+1=c\end{cases}}\)với a,b,c > 1, khi đó ta có
\(\hept{\begin{cases}ab=4\\bc=9\\ca=16\end{cases}}\Leftrightarrow\hept{\begin{cases}abbc=4.9\\c=\frac{9}{b}\\ca=16\end{cases}}\Leftrightarrow\hept{\begin{cases}16b^2=36\\c=\frac{9}{b}\\a=\frac{16}{c}\end{cases}}\Leftrightarrow\hept{\begin{cases}b^2=\frac{36}{16}=\frac{9}{4}\\c=\frac{9}{b}\\a=\frac{16}{c}\end{cases}}\Leftrightarrow\hept{\begin{cases}b=\frac{3}{2}\\c=\frac{9}{\frac{3}{2}}=6\\a=\frac{16}{6}=\frac{8}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=a-1=\frac{8}{3}-1=\frac{5}{3}\\y=b-1=\frac{3}{2}-1=\frac{1}{2}\\z=c-1=6-1=5\end{cases}}\)
Vậy \(P=x+y+z=\frac{5}{3}+\frac{1}{2}+5=\frac{10+3+30}{6}=\frac{43}{6}\)
\(f\left(x\right).g\left(x\right)+x^2.[1-3.g\left(x\right)]=\frac{5}{2}\)
\(\Rightarrow f\left(x\right).g\left(x\right)+x^2-3x^2.g\left(x\right)=\frac{5}{2}\) (1)
Ta thay: \(f\left(x\right)=3x^2-x+1\) và \(g\left(x\right)=x-1\) vào (1) ta được
\(\left(3x^2-x+1\right).\left(x-1\right)+x^2-3x^2.\left(x-1\right)=\frac{5}{2}\)
\(\Rightarrow\left(x-1\right).\left(3x^2-x+1-3x^2\right)+x^2=\frac{5}{2}\)
\(\Rightarrow\left(x-1\right).\left(-x+1\right)+x^2=\frac{5}{2}\)
\(\Rightarrow-\left(x-1\right)^2+x^2=\frac{5}{2}\)
\(\Rightarrow-x^2+2x-1+x^2=\frac{5}{2}\)
\(\Rightarrow2x=\frac{7}{2}\)
\(\Rightarrow x=\frac{7}{4}\)
a) \(f\left(x\right).g\left(x\right)=\left(3x^2-x+1\right).\left(x-1\right)\)
\(=x.\left(3x^2-x+1\right)-\left(3x^2-x+1\right)\)
\(=3x^3-x^2+x-3x^2+x-1\)
\(=3x^3-4x^2+2x-1\)
b) \(f\left(x\right).g\left(x\right)+x^2.\left[1-3.g\left(x\right)\right]=\frac{5}{2}\)
=> \(3x^3-4x^2+2x-1+x^2.\left(1-3x+3\right)=\frac{5}{2}\)
=> \(3x^3-4x^2+2x-1+x^2-3x^3+3x^2=\frac{5}{2}\)
=> \(2x-1=\frac{5}{2}\)
=>\(2x=\frac{5}{2}+1=\frac{5+2}{2}=\frac{7}{2}\)
=>\(x=\frac{7}{2}:2=\frac{7}{4}\)
\(9x^2+5y^2-6xy-6x-6y+20\)
\(=9x^2+y^2+1-6x+2y-6xy+4y^2-8y+4+15\)
\(=\left(3x-y-1\right)^2+4\left(y-1\right)^2+15\ge15\)
Dấu \(=\)khi \(\hept{\begin{cases}3x-y-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=1\end{cases}}\).
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Vì a chia cho 4 dư 2 nên đặt \(a=4k+2\left(k\inℕ\right)\)
\(\Rightarrow a^2=\left(4k+2\right)^2=16k^2+16k+4=4\left(4k^2+4k+1\right)⋮4\)
Vậy a2 chia cho 4 dư 0.