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Câu 1:
a) \(3,5+\sqrt{\frac{49}{25}}-\sqrt{0,36}\)
\(=3,5+\sqrt{1,96}-\sqrt{0,36}\)
\(=3,5+1,4-0,6\)
\(=4,9-0,6\)
\(=4,3.\)
Câu 2:
a) \(\frac{4}{9}:\left(x+0,4\right)=\frac{2}{3}\)
\(\Rightarrow\left(x+0,4\right)=\frac{4}{9}:\frac{2}{3}\)
\(\Rightarrow x+0,4=\frac{2}{3}\)
\(\Rightarrow x+\frac{2}{5}=\frac{2}{3}\)
\(\Rightarrow x=\frac{2}{3}-\frac{2}{5}\)
\(\Rightarrow x=\frac{4}{15}\)
Vậy \(x=\frac{4}{15}.\)
Bài 3:
Ta có: \(4x=5y.\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
\(\Rightarrow\frac{x}{5}=\frac{y}{4}\) và \(x+y=18.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{5}=\frac{y}{4}=\frac{x+y}{5+4}=\frac{18}{9}=2.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{5}=2\Rightarrow x=2.5=10\\\frac{y}{4}=2\Rightarrow y=2.4=8\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(10;8\right).\)
Chúc bạn học tốt!
Ta có:\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
\(\Rightarrow1+\frac{x+1}{11}+1+\frac{x+2}{10}=1+\frac{x+3}{9}+1+\frac{x+4}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}=\frac{x+12}{9}+\frac{x+12}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Mà \(\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)>0\)
\(\Rightarrow x+12=0\Rightarrow x=-12\)
\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
<=> \(\frac{x+1}{11}+\frac{x+2}{10}-\frac{x+3}{9}-\frac{x+4}{8}=0\)
<=> \(\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)-\left(\frac{x+3}{9}+1\right)-\left(\frac{x+4}{8}+1\right)=0\)<=> \(\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
<=> \(\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
<=> x + 12 = 0.Vì \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
<=> x = -12
a) Giải:
Ta có: \(a,b,c>0\Rightarrow a+b+c>0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2a+b}=\frac{a+b+c}{2b+c+2c+a+2a+b}=\frac{a+b+c}{3a+3b+3c}=\frac{a+b+c}{3\left(a+b+c\right)}=\frac{1}{3}\)
Vậy \(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2a+b}=\frac{1}{3}\)
Bài 1:
a)
\((\frac{3}{5})^2-[\frac{1}{3}:3-\sqrt{16}.(\frac{1}{2})^2]-(10.12-2014)^0\)
\(=\frac{9}{25}-(\frac{1}{9}-1)-1\)
\(=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\)
b)
\(|-\frac{100}{123}|:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})\)
\(=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}=(\frac{100}{123}+\frac{23}{123}):\frac{4}{3}=1:\frac{4}{3}=\frac{3}{4}\)
c)
\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)
\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)
Bài 2:
a)
\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b)
\((\frac{1}{2}-x)^2=(-2)^2=2^2\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)
c)
\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=\frac{2}{3}+1=\frac{5}{3}\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)
d)
\(3^{2x+1}=81=3^4\)
\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)
a)\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow x+\frac{1}{5}=2\) hoặc \(-2\)
Xét \(x+\frac{1}{5}=2\Leftrightarrow x=\frac{9}{5}\)
Xét \(x+\frac{1}{5}=-2\Leftrightarrow x=-\frac{11}{5}\)
\(\Leftrightarrow\dfrac{6}{5}< \dfrac{2x-3}{2}< \dfrac{12}{5}\)
=>12<5(2x-3)<24
\(\Leftrightarrow5\left(2x-3\right)\in\left\{15;20\right\}\)
\(\Leftrightarrow2x-3=3\)
hay x=3
a: Vì x/3=y/3 nên x=y
mà x+y=10
nên x=y=10/2=5
b: \(=\left(4+\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(\dfrac{5}{19}+\dfrac{14}{19}\right)+1.5=5.5+1=6.5\)
c: \(=9\cdot\dfrac{1}{3}-7+\left(-125\right):5=3-7-25=-29\)