Lời giải:
$\frac{1+5y}{5x}=\frac{1+7y}{4x}$
$\Rightarrow \frac{1+5y}{5}=\frac{1+7y}{4}$
$\Rightarrow 4(1+5y)=5(1+7y)$
$\Rightarrow 4+20y=5+35y$
$\Rightarrow -15y=1$
$\Rightarrow y=\frac{-1}{15}$
Lại có:
$\frac{1+5y}{5x}=\frac{1+3y}{12}$
$\Rightarrow 5x=(1+5y): \frac{1+3y}{12}=\frac{12(1+5y)}{1+3y}$
$\Rightarrow x=\frac{12(1+5y)}{5(1+3y)}$
Thay $y=\frac{-1}{15}$ thì $x=2$
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Rút gọn biểu thức
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)\)
Giúp em với ạ
Tính nhanh :
\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(-\dfrac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0.5\cdot\left(-2\dfrac{3}{5}\right)\)
1) Cho\(A=\frac{5\cdot x-4}{2\cdot x+5}-\frac{3\cdot y-3\cdot x}{2\cdot y-5}và3\cdot x-y=5\).Tính A
Bài 1: Tìm x,y:
\(\frac{2\cdot x+1}{5}=\frac{3\cdot y-2}{7}=\frac{2\cdot x+3\cdot y-1}{6\cdot x}\)
a)\(2\cdot\text{|}3-2x\text{|}+\dfrac{1}{2}=\dfrac{5}{2}\) b)\(x^2\cdot\left(2^x-6\right)-2x^3=0\)
Cho \(\dfrac{x}{3}\)=\(\dfrac{y}{4}\) và \(x\cdot y^2\)=384.Tìm x,y?
