=>\(\dfrac{140x+700-150x}{x\left(x+5\right)}=1\)
=>x^2+5x=-10x+700
=>x^2+15x-700=0
=>(x+35)(x-20)=0
=>x=20 hoặc x=-35
=>\(\dfrac{140x+700-150x}{x\left(x+5\right)}=1\)
=>x^2+5x=-10x+700
=>x^2+15x-700=0
=>(x+35)(x-20)=0
=>x=20 hoặc x=-35
giải pt
\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\)
Giải hệ phương trình sau
\(x+y=140\)
\(x-\frac{x}{8}=y+\frac{8}{x}\)
\(\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)
\(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(\left(\frac{x-2}{x+1}\right)^2+\frac{x^2-4}{x^2-1}-2\left(\frac{x+2}{x-1}\right)^2=0\)
\(x^2+\frac{x^2}{\left(x+1\right)^2}=3\)
\(\sqrt{x+3}-\sqrt{x-2}=1\)
\(\sqrt{5-x^2}+\sqrt{x^2+8}=5\)
1) Cho x,y>0 thỏa mãn x+y=1
Tìm Min A=\(\frac{1}{x^2+y^2}+\frac{1}{xy}\)
2) Giải phương trình
a) \(2\sqrt{x-1}+3\sqrt{5-x}=2\sqrt{13}\)
b) \(\sqrt{x+3}+\sqrt{x-1}=2\)
c) \(x^2-\sqrt{x+5}=5\)
Giúp mk bài này với
Giải các phương trình sau:
a, \(\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}=x\)
b, \(\sqrt{\frac{x^3+1}{x+3}}\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
c, \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
d, \(\sqrt{x^2-\frac{7}{x^2}}\sqrt{x-\frac{7}{x^2}}=x\)
e, \(\sqrt{\frac{5}{4}-x^2+\sqrt{1-x^2}}+\sqrt{\frac{5}{4}-x^2-\sqrt{1-x^2}}=x+1\)
g, \(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\)
h, \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
i, \(\sqrt[3]{x+34}-\sqrt[3]{x-3}=1\)
k, \(\sqrt[3]{3x-1}+\sqrt[3]{2x-1}=\sqrt[3]{5x+1}\)
giải phương trình
\(\sqrt{x}+\sqrt{1-x}+2\sqrt{x-x^2}-2\sqrt[4]{x-x^2}=1\)
\(\sqrt{x^2+10x+7}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x-3x+12}\)
\(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
\(x+4\sqrt{5-x}=4\sqrt{x-1}+\sqrt{-x^2+6x-5}+1\)
GIÚP MK VỚI
GIẢI HỆ PT
A;
\(\frac{5}{x-1}+\frac{1}{y-1}=10\)
\(\frac{1}{x-1}-\frac{3}{y-1}=18\)
B;
\(\frac{4}{x+2y}-\frac{1}{x-2y}=1\)
\(\frac{20}{x+2y}+\frac{3}{x-2y}=1\)
C;
\(\frac{12}{x-3}-\frac{5}{y+2}=63\)
\(\frac{8}{x-3}+\frac{15}{y+2}=-13\)
D;
\(\frac{5}{x+y-3}-\frac{2}{x-y+1}=8\)
\(\frac{3}{x+y-3}+\frac{1}{x-y+1}=1,5\)
Cho x (x thuộc R, x>0) thoả mãn \(x^2+\frac{1}{x^2}=7\)
a) Tính A = \(x^3+\frac{1}{x^3}\)
b) Tính B= \(x^5+\frac{1}{x^5}\)
Nhanh, mình dang gấp