HOC24
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\(\frac{8}{6x+2}\)=x-3
\(\Rightarrow\)(6x+2)(x-3)=8
\(\Rightarrow\)6x\(^2\)-18x+2x-6=8
\(\Rightarrow\)6x\(^2\)-18x+2x-6-8=0
\(\Rightarrow\)6x\(^2\)-16x-14=0
\(\frac{8}{6}\)x+2=x-3
\(\Leftrightarrow\)\(\frac{8}{6}\)x-x=-2-3
\(\Leftrightarrow\)\(\frac{1}{3}\)x=-5
\(\Leftrightarrow\)x=-5:\(\frac{1}{3}\)
\(\Leftrightarrow\)x=-15
Vậy x=-15
\(\frac{\sqrt{x}}{\sqrt{x}-2}-\left(\frac{2}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}\right)\)
ĐKXĐ:x\(\ge\)0 và x\(\ne\)4
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)-\(\left(\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)-\(\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)-2\left(\sqrt{x}-2\right)-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{x+2\sqrt{x}-2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}-2}{\sqrt{x}+2}\)(Thỏa mãn)