\(\Leftrightarrow3x+x^2-4x=12\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Delta=\left(-1\right)^2-4.\left(-12\right)\)
\(=1+48=49>0\)
=> pt có 2 nghiệm
\(\left\{{}\begin{matrix}x_1=\dfrac{1+\sqrt{49}}{2}=\dfrac{8}{2}=4\\x_2=\dfrac{1-\sqrt{49}}{2}=\dfrac{-6}{2}=-3\end{matrix}\right.\)
giải pt \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)=3x^2\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
a)\(\hept{\begin{cases}2\left(3x-2\right)-4=5\left(3y+2\right)\\4\left(3x-2\right)+7\left(3y+2\right)=-2\end{cases}}\)
b)\(\hept{\begin{cases}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=11\end{cases}}\)
Giúp mình nha
a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
giải phương trình bằng phương pháp đặt ẩn phụ:
ạ) \(2\sqrt{\left(-2x^2+5x+7\right)}=x^3-3x^2-x+12\)
b) \(x^2-3x+3=\left(4+3x-\frac{4}{x}\right)\sqrt{\left(x-1\right)}\)
Tìm x :\(\frac{15x}{\left(x-1\right)\left(x+4\right)}-1=12\left(\frac{1}{x+4}+\frac{1}{3x-3}\right)\)
Ai nhanh mình tick cho . Mk cảm ơn
Giaỉ các phương trình sau :
a) \(\frac{15x}{x^2+3x-4}=\frac{12}{x+4}+\frac{4}{x-1}+1\)
b) \(x\left(x-2\right)\left(x-1\right)\left(x+1\right)=24\)
Rút gọn:
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right).\sqrt{9-x^2}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right).\sqrt{x^2-6x+8}}\)
\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
Rút gọn:
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right).\sqrt{9-x^2}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right).\sqrt{x^2-6x+8}}\)
\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
