Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
D = \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\) . \(\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) = \(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
Câu hỏi của Phương Boice - Toán lớp 8 - Học toán với OnlineMath
Đặt \(\sqrt{x^2-x+1}=a\left(ĐK:a>0\right)\)
\(pt\Leftrightarrow\frac{\left(x^6+3x^4a\right)\left(4-a^2\right)}{4\left(2+a\right)a^2}=a\left(2-a\right)\)
\(\Leftrightarrow\left(x^6+3x^4a\right)\left(4-a^2\right)=4a^3\left(4-a^2\right)\)
\(\Leftrightarrow\left(4-a^2\right)\left(x^6+3x^4a-4a^3\right)=0\)
TH1: \(4-a^2=0\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=2\left(n\right)\end{cases}}\)
Với a = 2 , \(\sqrt{x^2-x+1}=2\Rightarrow x^2-x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}+1}{2}\\x=\frac{-\sqrt{13}+1}{2}\end{cases}}\)
TH2: \(x^6+3x^4a-4a^3=0\Rightarrow x^6-x^4a+4x^4a-4x^2a^2+4x^2a^2-4a^3=0\)
\(\Leftrightarrow\left(x^2-a\right)\left(x^4+4x^2a+4a^2\right)=0\Leftrightarrow\left(x^2-a\right)\left(x^2+2a\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=a\\x^2=-2a\left(l\right)\end{cases}}\)
Với \(x^2=a\Rightarrow x^2=\sqrt{x^2-x+1}\)
Đến đây bình phương và tìm ra nghiệm.
B= \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{x-2\sqrt{2}}\right):\left(\frac{1}{\sqrt{x}+2}+\frac{4}{x-4}\right)\)
= \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
= \(\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\):\(\frac{1}{\sqrt{x}-2}\)
=\(\frac{\sqrt{x}+2}{\sqrt{x}}.\sqrt{x}-2\)
Nhớ tick cho mk nhé
=\(\frac{x-4}{\sqrt{x}}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)
\(A=\frac{\sqrt{x}-2}{x-4}\)
\(\left(\frac{\sqrt{\left(-4\right).\left(-9\right)}}{\sqrt{2}}-\sqrt{2}.x\right):5,6=-7,2\)
\(\left(\frac{\sqrt{36}}{\sqrt{2}}-\sqrt{2}.x\right):\frac{28}{5}=\frac{-36}{5}\)
\(\frac{6}{\sqrt{2}}-\sqrt{2}.x=\frac{-36}{5}.\frac{28}{5}\)
\(\frac{6}{\sqrt{2}}-\sqrt{2}.x=\frac{-1008}{25}\)
\(\sqrt{2}.x=\frac{6}{\sqrt{2}}-\frac{-1008}{25}\)
\(\sqrt{2}.x=\frac{6}{\sqrt{2}}+\frac{1008}{25}\)
\(\sqrt{2}.x=\frac{150+\sqrt{2}.1008}{\sqrt{2}.25}\)
\(x=\frac{150+\sqrt{2}.1008}{\sqrt{2}.25}.\frac{1}{\sqrt{2}}\)
\(x=\frac{150+\sqrt{2}.1008}{25.2}=\frac{75+\sqrt{2}.504}{25}\)
Vậy \(x=\frac{75+\sqrt{2}.504}{25}\)
Khiếp bn quá ,toán lớp 8 cg giải dc