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1: Ta có: \(x^2-16=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
2: Ta có: \(1-36x^2=0\)
\(\Leftrightarrow\left(6x-1\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(7,=\left(0,5a+5b\right)\left(0,25a^2-2,5ab+25b^2\right)\\ 8,=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\\ 9,=\left(5+a-b\right)\left(25-5a+5b+a^2-2ab+b^2\right)\)
\(61,\\ 1,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ 2,\Leftrightarrow\left(1-6x\right)\left(1+6x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ 3,\Leftrightarrow\left(6-x\right)\left(6+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Bài 8:
a) \(\left(-3,5\right):\left(-2\dfrac{3}{5}\right)=\dfrac{7}{2}:\dfrac{13}{2}=\dfrac{7}{2}\cdot\dfrac{2}{13}=\dfrac{7\cdot2}{2\cdot13}=\dfrac{7}{13}\)
b) \(\left(-\dfrac{11}{15}\right):1\dfrac{1}{10}=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}=\left(-\dfrac{11}{15}\right)\cdot\dfrac{10}{11}=\dfrac{-11\cdot10}{15\cdot11}=-\dfrac{10}{15}=-\dfrac{2}{3}\)
c) \(2\dfrac{2}{3}:\left(-3\dfrac{3}{4}\right)=\dfrac{8}{3}:-\dfrac{15}{4}=\dfrac{8}{3}\cdot-\dfrac{4}{15}=\dfrac{8\cdot4}{3\cdot15}=-\dfrac{32}{45}\)
Bài 7:
a) \(\left(-\dfrac{3}{25}\right):6=\left(-\dfrac{3}{25}\right)\cdot\dfrac{1}{6}=\dfrac{-3\cdot1}{25\cdot6}=-\dfrac{1}{50}\)
b) \(-\dfrac{5}{23}:-2=\dfrac{5}{23}\cdot\dfrac{1}{2}=\dfrac{5\cdot1}{23\cdot2}=\dfrac{5}{26}\)
c) \(\dfrac{-7}{11}:-3,5=\dfrac{7}{11}:\dfrac{7}{2}=\dfrac{7}{11}\cdot\dfrac{2}{7}=\dfrac{7\cdot2}{11\cdot7}=\dfrac{2}{11}\)
a) \(2^4+8\left[\left(-2\right)^2:\dfrac{1}{2}\right]^0-2^{-2}.4+\left(-2\right)^2\)
\(=2^4+8.1-\dfrac{1}{4}.4+4\)
\(=16+8-1+4\)
\(=24-1+4\)
\(=23+4\)
\(=27\)
Bài 3:
a: Ta có: \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
\(=6x^2+9x+14x+21-6x^2-33x+10x+55\)
=76
b: Ta có: \(B=\left(x-3\right)\left(x+2\right)-\left(x-5\right)\left(x+4\right)\)
\(=x^2+2x-3x-6-x^2-4x+5x+20\)
=14
Câu 4:
Số đo các góc còn lại là \(47^0;133^0;133^0\)
Câu 6
a) Ta có: \(\widehat{A}=90^0\) ⇒a⊥c
a//b, a⊥c ⇒b⊥c
b) Ta lại có: M1+N1=1800(trong cùng phía)
1200+N1=1800
N1=1800-1200=600
cảm ơn ak