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\(\left\{{}\begin{matrix}x-y=7\\\left(x-y\right)\left(x^2+xy+y^2\right)=133\end{matrix}\right.< =>\left\{{}\begin{matrix}x-y=7\\x^2+xy+y^2=19\end{matrix}\right.< =>\left\{{}\begin{matrix}x-y=7\\\left(x-y\right)^2-xy=19\end{matrix}\right.< =>\left\{{}\begin{matrix}x-y=7\\xy=30\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7+y\\\left(7+y\right)y=30\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7+y\\y^2+7y-30=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7+y\\\left[{}\begin{matrix}y=3\\y=-10\end{matrix}\right.\end{matrix}\right.< =>\left\{{}\begin{matrix}\left[{}\begin{matrix}x=10\\x=-3\end{matrix}\right.\\\left[{}\begin{matrix}y=3\\y=-10\end{matrix}\right.\end{matrix}\right.\)Vậy....
\(\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\mx+2\left(3-x\right)=2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\mx-2x=2m-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\x\left(m-2\right)=2m-6\end{matrix}\right.\)
với \(\left\{{}\begin{matrix}m-2=0\\2m-6\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=2\\m\ne3\end{matrix}\right.\)
\(\Leftrightarrow m=2\)
Khi đó : \(\left\{{}\begin{matrix}x\in R\\y=3-x\end{matrix}\right.\)
⇔ hệ pt vô số nghiệm
\(m-2\ne0\)
\(\Leftrightarrow m\ne2\)
Khi đó hệ pt có nghiệm duy nhất là :
\(\left\{{}\begin{matrix}x=\frac{2m-6}{m-2}\\y=\frac{m}{m-2}\end{matrix}\right.\)
a)\(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)
b) \(\sqrt{75.48}=\sqrt{25.3.16.3}=\sqrt{5^2.3^2.4^2}=5.4.3=60\)
c)\(\sqrt{90.6,4}=\sqrt{10.9.4.1,6}=\sqrt{4^2.3^2.2^2}=4.3.2=24\)
d) \(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}}=\sqrt{\left(\dfrac{5.12}{10}\right)^2}=\dfrac{5.12}{10}=6\)
a) \(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)
b)\(\sqrt{75.48}=\sqrt{25.3.3.16}=5.3.4=60\)
c)\(\sqrt{90.6,4}=\sqrt{9.64}=3.8=24\)
d)\(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}=\dfrac{5.12}{10}=\dfrac{60}{10}=6}\)
a)\(\sqrt{75\cdot48}=\sqrt{25\cdot3\cdot48}=\sqrt{25\cdot144}=\sqrt{25}\cdot\sqrt{144}=5\cdot12=60\)
b) \(\sqrt{2,5\cdot14,4}=\sqrt{25\cdot144\cdot\frac{1}{100}}=\sqrt{25}\cdot\sqrt{144}\cdot\sqrt{\frac{1}{100}}=5\cdot12\cdot\frac{1}{10}=6\)
a) ĐS: 2.4.
b) ĐS: 28.
c) HD: Đổi 12,1.360 thành 121.36. ĐS: 66
d) ĐS: 18.
a) \(\sqrt{0,09.64}\)
\(=\sqrt{0,09}.\sqrt{64}\)
\(=0,3.8=2,4\)
b) \(\sqrt{2^4.\left(-7\right)^2}\)
\(=\sqrt{2^4}.\sqrt{\left(-7\right)^2}\)
\(=2^2.7=4.7=28\)
c) \(\sqrt{12,1.360}\)
\(=\sqrt{121.36}\)
\(=\sqrt{121}.\sqrt{36}\)
\(=11.6=66\)
d) \(\sqrt{2^2.3^4}\)
\(=\sqrt{2^2}.\sqrt{3^4}\)
\(=2.3^2=2.9=18\)
a) \(\sqrt{0,09.64}=\sqrt{\left(0,3\right)^2.8^2}=0,3.8=2,4\)
b) \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{\left(2^2\right)^2.\left(-7\right)^2}=2^2.\left|-7\right|=7.4=28\)
c) \(\sqrt{12,1.360}=\sqrt{12,1.10.36}=\sqrt{121.36}=\sqrt{11^2.6^2}=11.6=66\)
d) \(\sqrt{2^2.3^4}=\sqrt{2^2.\left(3^2\right)^2}=2.3^2=9.2=18\)
a) \(\sqrt{0,09\cdot64}=\sqrt{0,09}\cdot\sqrt{64}=0,3\cdot8=2,4\)
b) \(\sqrt{2^4\cdot\left(-7\right)^2}=\sqrt{2^4}\cdot\sqrt{\left(-7\right)^2}=2^2\cdot7=4\cdot7=28\)
c) \(\sqrt{12,1\cdot360}=\sqrt{12,1\cdot10\cdot36}=\sqrt{121\cdot36}=\sqrt{121}\cdot\sqrt{36}=11\cdot6=66\)
d) \(\sqrt{2^2\cdot3^4}=\sqrt{2^2}\cdot\sqrt{3^4}=2\cdot3^2=2\cdot9=18\)
\(\int ^{143242}_{43522523}dx=0\)
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