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a) \(10x-\frac{5}{18}+x+\frac{3}{12}=7x+\frac{3}{6}-12-\frac{x}{9}\)
\(10x+x-7x+\frac{x}{9}=\frac{3}{6}+\frac{5}{18}-\frac{3}{12}-12\)
\(\frac{37}{9}x=-\frac{413}{36}\)
\(x=-\frac{413}{148}\)


17) \(\left(x^2-11x+30\right)\left(x^2-13x+30\right)=24x^2\)
\(\left(x-11+\frac{30}{x}\right)\left(x-13+\frac{30}{x}\right)=24\)
\(t\left(t-2\right)=24\)
\(\left(t-1\right)^2=25\)
t =6 hoặc t =-4
+\(\left(x-11+\frac{30}{x}\right)=6\Leftrightarrow x^2-11x+30=6x\Leftrightarrow x^2-17x+30=0\)
+\(\left(x-11+\frac{30}{x}\right)=-4\)

C=(x-3)^2 -9 >= 9
D=(2x-1)^1 -6>=6
E=2(x-3)^2 -21>=21
F=....................

1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$
ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{6}{x-2}+\dfrac{5}{x+2}=\dfrac{x-18}{x^2-4}\\ \Leftrightarrow\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x-18}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{6x+12+5x-10-x+18}{\left(x+2\right)\left(x-2\right)}=0\\ \Rightarrow10x+20=0\\ \Leftrightarrow x=-2\left(ktm\right)\)
\(\dfrac{6}{x-2}-\dfrac{5}{x+2}=\dfrac{x-18}{x}\)
mũ 4 ở đâu vậy ạ