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a/\(\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(\Rightarrow1-x^2=\frac{7}{16}\)
\(\Rightarrow x^2=\frac{9}{16}\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{3}{4}\\-\frac{3}{4}\end{cases}}\)
\(a,\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(1-x^2=\frac{7}{16}\)
\(x^2=1-\frac{7}{16}=\frac{9}{16}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(b,x^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(x^2+-\frac{9}{25}=\frac{16}{25}\)
\(x^2=\frac{16}{25}--\frac{9}{25}=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
học tốt ~~~
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
Nên x + 1 = 0 => x = -1
b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)
Nên x +15 = 0 => x = -15
a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1
a, (x2 - 5)(x2 - 24) < 0
=> x2 - 5 và x2 - 24 trái dấu
Mà x2 - 5 > x2 - 24 => \(\hept{\begin{cases}x^2-5>0\\x^2-24>0\end{cases}\Rightarrow5< x^2< 24}\)
Vì x \(\in\)Z nên x2 = 9;16
+) x2 = 9 => x = 3 hoặc x = -3
+) x2 = 16 => x = 4 hoặc x = -4
Vậy...
b,
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
=> x + 1 = 0 => x = 0 - 1 => x = -1
\(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Rightarrow\left(\frac{x+1}{14}+1\right)+\left(\frac{x+2}{13}+1\right)=\left(\frac{x+3}{12}+1\right)+\left(\frac{x+4}{11}+1\right)\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Rightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Mà \(\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)\ne0\)
=> x + 15 = 0 => x = 0 - 15 => x = -15
a,\(\frac{-2}{3}+\frac{1}{5}=\frac{-2.5}{3.5}+\frac{3.1}{3.5}=-\frac{10}{15}+\frac{3}{15}=-\frac{7}{15}\)
b,\(3\frac{11}{13}-5\frac{11}{3}=3+\frac{11}{13}-(5+\frac{11}{13})=\left(3-5\right)+\left(\frac{11}{13}-\frac{11}{13}\right)=-2+0=-2\)
c,\(1\frac{2}{3}:(-\frac{5}{3})=\frac{5}{3}:\left(-\frac{5}{3}\right)=\frac{5.3}{3.\left(-5\right)}=\frac{15}{-15}=-1\)
d,\(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)=1+\left(-1\right)=0\)
Tìm x:
a,x+12=8
x=8-12
x=-4
b,\(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{5}{10}\)
\(\frac{2}{3}x=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{2}{3}\)
\(x=\frac{6}{10}\)
\(x=\frac{3}{5}\)
a) \(\frac{-2}{3}+\frac{1}{5}=\frac{-10}{15}+\frac{3}{15}=\frac{-10+3}{15}=\frac{-7}{15}\)
b) \(3\frac{11}{13}-5\frac{11}{13}=\frac{50}{13}-\frac{76}{13}=\frac{50-76}{13}=\frac{-26}{13}=-2\)
c) \(1\frac{2}{3}:\frac{-5}{3}=\frac{5}{3}:\frac{-5}{3}=\frac{5}{3}\times\frac{3}{-5}=\frac{15}{-15}=-1\)
d) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)
\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)
\(=\left(\frac{31-14}{17}\right)+\left(\frac{-5-8}{13}\right)\)
\(=\frac{17}{17}+\frac{-13}{13}\)
\(=1+\left(-1\right)\)
\(=0\)
TÌM x
a) \(x+12=8\)
\(\Leftrightarrow x=8-12\)
\(\Leftrightarrow x=-4\)
b) \(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{1-5}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{-4}{10}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}:\frac{2}{3}=\frac{-2}{5}\times\frac{3}{2}\)
\(\Leftrightarrow x=\frac{-6}{10}=\frac{-3}{5}\)
3) Giải:
Ta có bảng sau:
x - 2 | 1 | 11 |
y - 1 | 11 | 1 |
x | 3 | 13 |
y | 12 | 2 |
Vậy cặp số \(\left(x;y\right)\) là \(\left(3;12\right);\left(12;3\right)\)
4) Giải:
Ta có bảng sau:
x - 3 | 13 | 1 |
y - 2 | 1 | 13 |
x | 16 | 4 |
y | 3 | 15 |
Vậy cặp số \(\left(x;y\right)\) là \(\left(16;3\right);\left(4;15\right)\)
a ) x = 21 55 ;
b ) x = − 11 26