(x - 1/4)^3 = - 27^125
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\(\Rightarrow-5\left(n+3\right)+42⋮n+3\\ \Rightarrow n+3\inƯ\left(42\right)=\left\{-42;-21;-14;-7;-6;-3;-2;-1;1;2;3;6;7;14;21;42\right\}\\ \Rightarrow n\in\left\{-45;-24;-17;-10;-9;-6;-5;-4;-2;-1;0;3;4;11;17;39\right\}\)
1,
\(\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}\) và \(x^4.y^4=81\)
Đặt \(x^2=a\left(a\ge0\right);y^2=b\left(b\ge0\right)\)
Ta có \(\frac{a+b}{10}=\frac{a-2b}{7}\)và \(a^2b^2=81\)
:\(\frac{a+b}{10}=\frac{a-2b}{7}=\frac{\left(a+b\right)-\left(a-2b\right)}{10-7}=\frac{3b}{3}=b\) (1)
\(\frac{a+b}{10}=\frac{a-2b}{7}=\frac{2a+2b}{20}=\frac{\left(2a+2b\right)+\left(a-2b\right)}{20+7}=\frac{3a}{27}=\frac{a}{9}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{9}=b\Rightarrow a=9b\)
Do \(a^2b^2=81\)nên \(\left(9b^2\right).b^2=81\Rightarrow81b^4=81\Rightarrow b^4=1\Rightarrow b=1\left(b\ge0\right)\)
Suy ra a = 9 . 1 = 9
Ta có x2 = 9 và y2 = 1. Suy ra x = ±3, y = ±1.
\(x^4y^4=81\Rightarrow x^2y^2=9\Rightarrow x^2=\frac{9}{y^2}\)
\(\Rightarrow\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}\Leftrightarrow\frac{y^4+9}{10y^2}=\frac{9-2y^4}{7y^2}\Leftrightarrow7\left(y^4+9\right)=10\left(9-2y^4\right)\Leftrightarrow y^4=1\Leftrightarrow y=\pm1\)
\(\Rightarrow x^4=81\Leftrightarrow x=\pm3\)
x | y | 2012x+2013y |
1 | 3 | 8051 |
1 | -3 | -4027 |
-1 | 3 | 4027 |
-1 | -3 | -8051 |
\(x-\frac{3}{5}=1\div3\)
\(x-\frac{3}{5}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{3}{5}=\frac{14}{15}\)
Chắc x 1/6 = 5 : 4 là phép nhân
\(x\times\frac{1}{6}=5\div4\)
\(x\times\frac{1}{6}=\frac{5}{4}\)
\(x=\frac{5}{4}\div\frac{1}{6}\)
\(x=\frac{15}{2}\)
x-3/5=1/3
x=1/3+3/5
x=14/15
x.1/6=5/4
x=5/4:1/6
x=5/4x6
x=15/2
\(a.\frac{19}{5}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{19}{5}-\frac{4}{5}\)
\(=\frac{19}{5}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)-\frac{4}{5}\)
\(=\frac{19}{5}\cdot1-\frac{4}{5}\)
\(=\frac{19}{5}-\frac{4}{5}=\frac{15}{5}=3\)
\(b.2\frac{2}{7}\cdot5\frac{2}{5}+\frac{16}{7}\cdot1\frac{3}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\frac{27}{5}+\frac{16}{7}\cdot\frac{8}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\left(\frac{27}{5}+\frac{8}{5}\right)+\frac{1}{2}\)
\(=\frac{16}{7}\cdot7+\frac{1}{2}\)
\(=16+\frac{1}{2}=\frac{33}{2}\)
\(c.\frac{3}{7}\cdot3\frac{3}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{15}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\left(\frac{15}{4}-\frac{5}{4}\right)-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{5}{2}-\frac{1}{4}\)
\(=\frac{15}{14}-\frac{1}{4}=\frac{23}{28}\)
Chú ý: \(\cdot:\times\)
a) (2x - 5)3 = 27
(2x - 5)3 = 33
2x - 5 = 3
2x = 8
x = 4
b) 5x + 3 - 13 = 612
5x + 3 = 625
5x + 3 = 54
x + 3 = 4
x = 1
mk giải 1 câu thôi nhá,, đang bận
1. - 20 < x< 21
-> x\(\varepsilon\) I-19, -18, -17,..... 20I
tổng dãy tên là
-19+ -18+-17+...+20= 20
bn tk mk thì mk lm thêm cho
Ta có: \(\left(\frac{-3}{5}\right)^3=\frac{-27}{125}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^3=\left(\frac{-3}{5}\right)^3\)
\(\Rightarrow x-\frac{1}{4}=\frac{-3}{5}\)
\(\Rightarrow x=\frac{-7}{20}\)