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4 tháng 3 2020

mình sửa lại bài 3 ý a, \(\left|5x-3\right|< 2\)

19 tháng 3 2020

Giúp mình hoàn thành các bài tập này với ạ.Cảm ơn rất nhìuuuuuuu @@@

19 tháng 3 2020

@Akai Haruma

giải các hệ BPT sau: a) \(\left\{{}\begin{matrix}5x-24x+5\\5x-4< x+2\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\) e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\) f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\) g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\) h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\) j)...
Đọc tiếp

giải các hệ BPT sau:

a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)

e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)

f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)

g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)

h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)

j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)

3
25 tháng 3 2020
https://i.imgur.com/NOxfqjV.jpg
25 tháng 3 2020
https://i.imgur.com/awOKwJi.jpg
NV
27 tháng 10 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

NV
27 tháng 10 2019

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)