Giải các bpt sau
a, \(\frac{\left(4-x\right)\left(x^2-2x-15\right)}{2x^2+x+1}\le0\)
b, \(\frac{x^2+x-3}{x^2-4}\ge1\)
1. giải các hệ bất phương trình sau
a. \(\left\{{}\begin{matrix}\frac{2x+3}{x-1}\ge1\\\frac{\left(x+2\right)\left(2x-4\right)}{x-1}\le0\end{matrix}\right.\)
Giải bất phương trình
1) \(\frac{x^4-1}{x^2+3x}+x^2\ge1\)
2) \(\left(x^4-5x^2+4\right)\left(\frac{x-2}{x}-3\right)\le0\)
3) \(\left(\frac{4}{x}-\frac{2}{x-1}\right)\left(\frac{x^2+1}{x}-2\right)\le0\)
4) \(\left(\sqrt{x^3-4x}-\sqrt{15}\right)\sqrt{\frac{1+x}{x}-2}\le0\)
a/
\(\Leftrightarrow\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+3x}+x^2-1\ge0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{x^2+1}{x^2+3x}+1\right)\ge0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{2x^2+3x+1}{x^2+3x}\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(2x+1\right)}{x\left(x+3\right)}\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+1\right)\left(x+1\right)^2}{x\left(x+3\right)}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x< -3\\x=-1\\-\frac{1}{2}\le x< 0\\x\ge1\end{matrix}\right.\)
b/
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\left(\frac{-2-2x}{x}\right)\le0\)
\(\Leftrightarrow\frac{-2.\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}{x}\le0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x+1\right)^2}{x}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\le-2\\x=-1\\0< x\le1\\x\ge2\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(\frac{4\left(x-1\right)-2x}{x\left(x-1\right)}\right)\left(\frac{x^2+1-2x}{x}\right)\le0\)
\(\Leftrightarrow\frac{\left(2x-4\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)
\(\Rightarrow1< x\le2\)
d/
ĐKXĐ: \(\left\{{}\begin{matrix}x^3-4x\ge0\\\frac{1+x}{x}-2\ge0\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)\left(x+2\right)\ge0\\\frac{1-x}{x}\ge0\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}-2\le x\le0\\x\ge2\end{matrix}\right.\\0< x\le1\\x\ne0\end{matrix}\right.\)
\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ
Vậy BPT đã cho vô nghiệm
Giải các bất phương trình sau
a \(\frac{x^3-2x^2+4x}{-x^2+x+12}>0\)
b \(\frac{4x-3}{x-2}>7-\frac{3x-4}{x+3}\)
c \(\frac{\left(3-x\right)\left(x^2-4x+4\right)}{x^3-x}\le0\)
d \(\frac{2x-3}{3x+5}< \frac{3x+5}{2x-3}\)
e \(\frac{3x+2}{\left(x+1\right)\left(x+2\right)}\ge1\)
f \(\frac{x^3-3}{x^2-1}\ge3\)
Giải bpt sau:
\(\frac{\left(x-1\right)^3\left(x+2\right)^4\left(x-3\right)^5\left(x+6\right)}{x^2\left(x-7\right)^3}\le0\)
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
bài 1giải bpt
a) \(\frac{x+2}{3}-x+1>x+3\)
b) \(\frac{3x+5}{2}-1\le\frac{x+2}{3}+x\)
c) \(\frac{\left(x-2\right)\sqrt{x-1}}{\sqrt{x-1}}< 2\)
bài 2 \ giải hệ bpt
a) \(\left\{{}\begin{matrix}2-x>0\\2x+1>x-2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\frac{2x-1}{3}< -x+1\\\frac{4-3x}{2}< 3-x\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}-2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
Mgọi người giúp mình với ạ
Giải các bất phương trình sau:
a. \(\left(1-2x\right)\left(x^2+x-30\right)\left(x^2-4x+4\right)\le0\)
b. \(\frac{1}{x^2-5x+4}< \frac{1}{x^2-7x+6}\)
c. \(\frac{2}{x^2-x+1}-\frac{1}{x-1}\ge\frac{2x-1}{x^3+a}\)
Giúp mình với ạ
Giải các phương trình sau:
a)\(\frac{\left(9x-0.7\right)}{4}-\frac{\left(5x-1.5\right)}{7}=\frac{\left(7x-1.1\right)}{3}-\frac{5\left(0.4-2x\right)}{6}\)
b)\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
c)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)
d)\(\frac{8x^2}{3\left(1-4x\right)^2}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
Đang cần gấp. Ai nhanh+đúng 3tiks
Giải các pt sau
\(a,\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{3}\left(x+2\right)\)
\(b,\left(2x+1\right)^2=\left(x-1\right)^2\)
\(c,\left(x^2-5\right)\left(x+3\right)=0\)
\(d,\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
\(e,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(a,\frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}\)
\(\frac{13}{12}x=\frac{13}{12}\Rightarrow x=1\)
\(b,\left(2x+1\right)^2=\left(x-1\right)^2\Rightarrow\orbr{\begin{cases}2x+1=x-1\\2x+1=1-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)
\(c,\left(x^2-5\right)\left(x+3\right)=0\Rightarrow\left(x+5\right)\left(x-5\right)\left(x+3\right)=0\)
\(\Rightarrow x=\left\{-3;-5;5\right\}\)