Giải pt: x+3 – $\sqrt[]{14x-15}$ = (1-$\sqrt[]{10x-19}$ ):(1-x) 10/08/2021 Bởi Sadie Giải pt: x+3 – $\sqrt[]{14x-15}$ = (1-$\sqrt[]{10x-19}$ ):(1-x)
Giải thích các bước giải: $x+3-\sqrt{14x-15}=\dfrac{1-\sqrt{10x-19}}{1-x}$ $\to (x+3-\sqrt{14x-15})(1-x)=1-\sqrt{10x-19}$ $\to x^2+2x-x\sqrt{14x-15}+\sqrt{14x-15}-\sqrt{10x-19}-2=0$ $\to x(x+2-\sqrt{14x-15})-((x+2)-\sqrt{14x-15})+(x-\sqrt{10x-19})=0$ $\to x.\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}-\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}+\dfrac{x^2-(10x-19)}{x+\sqrt{10x-19}}=0$ $\to x.\dfrac{x^2-10x+19}{x+2+\sqrt{14x-15}}-\dfrac{x^2-10x+19}{x+2+\sqrt{14x-15}}+\dfrac{x^2-10x+19}{x+\sqrt{10x-19}}=0$ $\to x^2-10x+19=0\to x=5+\sqrt{6},\:x=5-\sqrt{6}$ Hoặc $ \dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}=0(*)$ Vì $x>\dfrac{19}{10}\to x+2+\sqrt{14x-15}>x+\sqrt{10x-19}$ $\to \dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}>0$ $\to (*)$ vô nghiệm Bình luận
Giải thích các bước giải:
$x+3-\sqrt{14x-15}=\dfrac{1-\sqrt{10x-19}}{1-x}$
$\to (x+3-\sqrt{14x-15})(1-x)=1-\sqrt{10x-19}$
$\to x^2+2x-x\sqrt{14x-15}+\sqrt{14x-15}-\sqrt{10x-19}-2=0$
$\to x(x+2-\sqrt{14x-15})-((x+2)-\sqrt{14x-15})+(x-\sqrt{10x-19})=0$
$\to x.\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}-\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}+\dfrac{x^2-(10x-19)}{x+\sqrt{10x-19}}=0$
$\to x.\dfrac{x^2-10x+19}{x+2+\sqrt{14x-15}}-\dfrac{x^2-10x+19}{x+2+\sqrt{14x-15}}+\dfrac{x^2-10x+19}{x+\sqrt{10x-19}}=0$
$\to x^2-10x+19=0\to x=5+\sqrt{6},\:x=5-\sqrt{6}$
Hoặc $ \dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}=0(*)$
Vì $x>\dfrac{19}{10}\to x+2+\sqrt{14x-15}>x+\sqrt{10x-19}$
$\to \dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}>0$
$\to (*)$ vô nghiệm