Lời giải:
ĐKXĐ: $x\geq 5$
PT$\Leftrightarrow \sqrt{x-5}=7-x$
\(\Rightarrow \left\{\begin{matrix} 7-x\geq 0\\ x-5=(7-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 7\\ x^2-15x+54=0\end{matrix}\right.\left\{\begin{matrix} x\leq 7\\ (x-9)(x-6)=0\end{matrix}\right.\)
$\Rightarrow x=6$ (tm)
Giải phương trình :
\(4^{x^2-3x+2}+4^{x^2+6x+5}=4^{2x^2+3x+7}+1\)
Bài 1
a,(2x+1)2/5 - (x-1)2/3=7x2-14x-5/15
b,(x+10)(x+4)/12 - (x-4)(2-x)/4=(x+10)(x-2)/3
c,(x-2)2/3 - (2x-3)(2x+3)/8 + (x-4)2/6=0
Bài 2
a,x-23/24 + x-23/25=x-23/26 + x-23/27
b,(x+2/98 + 1)+(x+3/97 + 1)=(x+4/96 + 1)+(x+5/95 + 1)
c,x+1/2004 + x+2/2003=x+3/2002 + x+4/2001
d,201-x/99 + 203-x/97=205-x/95 + 3=0
e,x-45/55 + x-47/53=x-55/45 + x-53/47
f,x+1/9 + x+2/8=x+3/7 + x+4/6
g,x+2/98 +x+4/96=x+6/94 + x+8/92
h,2-x/2002 - 1=1-x/2003 - x/2004
i,x2-10x-29/1971 + x2-10x -27/1973=x2-10x-1971/29 + x2-10x-1973/27
Hơi khó hiểu thông cảm ạ
\(\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{\left(7+x\right)^2}-\sqrt[3]{\left(7+x\right)\left(2-x\right)}=3\)
2x-1/x+1 +3x+1/x+2=x-7/x-1 +4
\(\dfrac{2x-1}{x+1}+\dfrac{3x-1}{x+2}=\dfrac{x-7}{x-1}+4\)
CHUYÊN ĐỀ GIẢI PHƯƠNG TRÌNH
a, \(\sqrt{2x-1}+\sqrt{x^2+3}=4-x\) f, \(2x^2-11x+23=4\sqrt{x+1}\)
b, \(\sqrt{x^2+x+1}=\sqrt{x^2-3x-1}+2x+1\) g, \(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
c, \(\left|x-16\right|^4+\left|x-17\right|^3=1\) h, \(9\left(\sqrt{4x+1}-\sqrt{3x-2}\right)=x+3\)
d, \(\left(x+1\right)\sqrt{x+2}+\left(x+6\right)\sqrt{x+7}=x^2+7x+12\)
e, \(\left(4x^3-x+3\right)^3-x^3=\frac{3}{2}\)
1.\(\sqrt{5x-1}-\sqrt{3x-2}-\sqrt{x-1}=0 \)
2.\(\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{\left(7+x\right)^2}-\sqrt[3]{\left(2-x\right)\left(7+x\right)}=3\)
giúp em đi mà =.=~
Giải các hệ phương trình sau :
a, \(\begin{cases}5x-4y=3\\7x-9y=8\end{cases}\)
b, \(\begin{cases}\frac{1}{x}-\frac{8}{y}=18\\\frac{5}{x}+\frac{4}{y}=51\end{cases}\)
c, \(\begin{cases}\frac{10}{x-1}+\frac{1}{y+2}=1\\\frac{25}{x-1}+\frac{3}{y+2}=2\end{cases}\)
d, \(\begin{cases}\frac{27}{2x-y}+\frac{32}{x+3y}=7\\\frac{45}{2x-y}-\frac{48}{x+3y}=-1\end{cases}\)
Giải phương trình :
\(4^{x+1}+4^{x-1}-2^{x+2}-2^{2-x}-7=0\)