Решите уравнение:
\(1)\ \frac{2x+7}{4}=\frac{x+5}{3};\)
\(2)\ x^2+6x=0;\)
\(3)\ 0{,}21x-0{,}7x^2=0;\)
\(\vphantom{\frac{0}{0}}4)\ x^2-16=0;\)
\(5)\ 25x^2-36=0;\)
\(6)\ x^2+4=0.\)
Упражнение 201
Источник заимствования: Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций / А.Г. Мерзляк, В.Б. Полонский, М.С. Якир – Вентана-Граф, 2019. – 47 c. ISBN 978-5-360-07402-1
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Решение:
\(1)\ \frac{2x+7}{4}=\frac{x+5}{3}\ | \cdot\ 12\)
\(\phantom{1)\ }3(2x+7)=4(x+5)\)
\(\phantom{1)\ }6x+21=4x+20\)
\(\phantom{1)\ }6x-4x=20-21\)
\(\phantom{1)\ }2x={-}1\)
\(\phantom{1)\ }x={-}\frac{1}{2}\)
\(\phantom{1)\ }3(2x+7)=4(x+5)\)
\(\phantom{1)\ }6x+21=4x+20\)
\(\phantom{1)\ }6x-4x=20-21\)
\(\phantom{1)\ }2x={-}1\)
\(\phantom{1)\ }x={-}\frac{1}{2}\)
\(2)\ x^2+6x=0\)
\(\phantom{2)\ }x(x+6)=0\)
\(\phantom{2)\ }\begin{array}[t]{lcl}x=0&\largeили&x+6=0\\[-1ex]\\&&x={-}6\end{array}\)
\(\phantom{2)\ }x(x+6)=0\)
\(\phantom{2)\ }\begin{array}[t]{lcl}x=0&\largeили&x+6=0\\[-1ex]\\&&x={-}6\end{array}\)
\(3)\ 0{,}21x-0{,}7x^2=0\)
\(\phantom{3)\ }0{,}7x(0{,}3-x)=0\)
\(\phantom{3)\ }\begin{array}[t]{lcl}x=0&\largeили&0{,}3-x=0\\[-1ex]\\&&x=0{,}3\end{array}\)
\(\phantom{3)\ }0{,}7x(0{,}3-x)=0\)
\(\phantom{3)\ }\begin{array}[t]{lcl}x=0&\largeили&0{,}3-x=0\\[-1ex]\\&&x=0{,}3\end{array}\)
\(\vphantom{\frac{0}{0}}4)\ x^2-16=0\)
\(\phantom{4)\ }x^2=16\)
\(\phantom{4)\ }\begin{array}[t]{lcl}x=4&\largeили&x={-}4\end{array}\)
\(\phantom{4)\ }x^2=16\)
\(\phantom{4)\ }\begin{array}[t]{lcl}x=4&\largeили&x={-}4\end{array}\)
\(5)\ 25x^2-36=0\)
\(\phantom{5)\ }25x^2=36\)
\(\phantom{5)\ }x^2=\frac{36}{25}\)
\(\phantom{5)\ }\begin{array}[t]{lcl}x=\frac{6}{5}&\largeили&x={-}\frac{6}{5}\\[-1ex]\\x=1\frac{1}{5}&&x={-}1\frac{1}{5}\end{array}\)
\(\phantom{5)\ }25x^2=36\)
\(\phantom{5)\ }x^2=\frac{36}{25}\)
\(\phantom{5)\ }\begin{array}[t]{lcl}x=\frac{6}{5}&\largeили&x={-}\frac{6}{5}\\[-1ex]\\x=1\frac{1}{5}&&x={-}1\frac{1}{5}\end{array}\)
\(6)\ x^2+4=0\)
\(\phantom{6)\ }x^2={-}4\)
\(\phantom{6)}\largeкорней\ нет\)
\(\phantom{6)\ }x^2={-}4\)
\(\phantom{6)}\largeкорней\ нет\)