\(ĐKXĐ:x\ne\frac{1}{5},x\ne\frac{3}{5}\)
Ta có : \(\frac{3}{5x-1}=\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\frac{3\left(3-5x\right)}{\left(5x-1\right)\left(3-5x\right)}-\frac{2\left(5x-1\right)}{\left(5x-1\right)\left(3-5x\right)}+\frac{4}{\left(5x-1\right)\left(5x-3\right)}=0\)
\(\Rightarrow9-15x-10x+2+4=0\)
\(\Leftrightarrow-25x=-15\)
\(\Leftrightarrow x=\frac{3}{5}\) ( không thỏa mãn \(ĐKXĐ\) )
Vậy pt đã cho vô nghiệm
Cho \(x=\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
Tính \(A=\left(4x^5+4x^4-x^3+1\right)^{19}+\left(\sqrt{x^5+4x^4-5x^3+5x+3}\right)^3+\left(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\right)\)
gpt:
\(3\left(x^2-3x+1\right)+\sqrt{3\left(x^4+x^2+1\right)}=0\)
\(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)
giải hệ phương trình:
1) \(\hept{\begin{cases}2\left(x+y\right)+3\left(x+y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{cases}}\)
2)\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12_{ }\end{cases}}\)
3) \(\hept{\begin{cases}\frac{2y-5x}{3}+5=\frac{y+27}{4}-2x\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{cases}}\)
4)\(\hept{\begin{cases}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{cases}}\)
5)\(\hept{\begin{cases}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{cases}}\)
tìm đa thức P(x) bậc 3 sao cho P(x) chia cho \(\left(X^2-5X+4\right)\)được dư là \(\left(\frac{X}{3}-\frac{2}{5}\right)\).Chia cho \(\left(X^2-5X+6\right)\)được dư là \(\left(\frac{X}{5}-\frac{2}{3}\right)\)
tính
\(\left(4x^3+4x^4-5x^3+5x-2\right)^{2015}+2016\)
với x=\(\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
giai phuong trinh a) \(\left(\frac{x+3}{x+2}\right)^2+6\cdot\left(\frac{x-3}{x-2}\right)^2-5\cdot\frac{x^2-9}{x^2-4}=0\)
b)(12x-1)*(x-1)*(4x-1)*(3x-1)=330
c)\(\frac{4x}{x^2+5x+3}+\frac{5x}{x^2-5x+3}=\frac{-3}{2}\)
Giải phương trình
\(\left(x+3\right)\sqrt{\left(4-x\right)\left(12+x\right)}=28-x\)
\(\sqrt{3x^3-5x^2+5x+5x-2}=\frac{x^2}{2}+x-\frac{1}{2}\)
\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
Bài 1 : Cho biểu thức R = \(\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\cdot\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ Rút gọn R
b/ Tìm các giá trị của x để R < -1
Bài 2 : Cho \(\sqrt{x^2-5x+14}-\sqrt{x^2-5x+10}=2\)Tính giá trị biểu thức M =\(\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}\)
Bài 3 : Tìm GTNN của : Q = \(\sqrt{x^2+4x+4}+\sqrt{x^2-4x+4}\)
Giải pt:
1) \(\frac{5x}{2x+3}\)+ 6 . \(\frac{5x+3}{x}\)= -3
2) \(\left(\frac{x+1}{x}\right)^2\)- 4.\(\left(\frac{x+1}{x}\right)\)+ 3 = 0
3) ( \(x^2\)+ \(3x-4\)).( \(x^2\)+\(x-6\)) = 24
4) \(\left(5x+4\right)^2\).\(\left(5x+8x\right)\)= 16
