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\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)
Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)
\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)
Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)
\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)
Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)
c) \(8x^3-1=8x^2+4x+2\)
<=> \(\left(2x-3\right)\left(4x^2+2x+1\right)=0\)
<=> \(2x-3=0\) hoặc \(4x^2+2x+1=0\)
Th1: x=\(\dfrac{3}{2}\)
Th2: Vô nghiệm
Vậy x=\(\dfrac{3}{2}\)
\(\text{a) }\dfrac{2x^2-x-1}{2}-3x^2+x+4=\left(5-x\right)\left(2x+4\right)\\ \Leftrightarrow\left(\dfrac{2x^2-x-1}{2}-3x^2+x+4\right)2=\left(5-x\right)\left(2x+4\right)2\\ \Leftrightarrow2x^2-x-1-6x^2+2x+8=\left(5-x\right)\left(4x+8\right)\\ \Leftrightarrow-4x^2+x+7=20x+40-4x^2-8x\\ \Leftrightarrow-4x^2+x+4x^2-12x=40-7\\ \Leftrightarrow-11x=33\\ \Leftrightarrow x=-3\\ \text{Vậy }S=\left\{-3\right\}\)
\(\text{b) }\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\dfrac{\left(x-1\right)\left(2x+4\right)}{2}+1\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\left(x-1\right)\left(x+2\right)+1\\ \Leftrightarrow\left(\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1\right)4=\left(x^2-x+2x-2+1\right)4\\ \Leftrightarrow\left(2x-5\right)\left(3x+7\right)+8x-4=\left(x^2+x-1\right)4\\ \Leftrightarrow6x^2-15x+14x-35+8x-4=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-39=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-4x^2-4x-39+4=0\\ \Leftrightarrow2x^2+3x-35=0\\ \Leftrightarrow2x^2+10x-7x-35=0\\ \Leftrightarrow\left(2x^2+10x\right)-\left(7x+35\right)=0\\ \Leftrightarrow2x\left(x+5\right)-7\left(x+5\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-5\end{matrix}\right.\\ \\ \text{Vậy }S=\left\{\dfrac{7}{2};-5\right\}\)
\(\text{c) }8x^3-1=8x^2+4x+2\\ \Leftrightarrow\left(2x-1\right)\left(4x^2+2x+1\right)=2\left(4x^2+2x+1\right)\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }S=\left\{\dfrac{3}{2}\right\}\)
\(\text{d) }\left(x^2+x+1\right)\left(x^2-x+1\right)=x^6-1\\ \Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-1\right)=1\\ \Leftrightarrow x^2-1=1\\ \Leftrightarrow x^2=2\\ \Leftrightarrow x=\sqrt{2}\\ \text{Vậy }S=\left\{\sqrt{2}\right\}\)
\(\text{e) }\left(x^3+2x\right)\left(x^2+4\right)=\left(x^2+6x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+2x^2+4x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[\left(x^2+2x^2\right)+\left(4x^2+8\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[x^2\left(x^2+2\right)+4\left(x^2+2\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+4\right)\left(x^2+2\right)\left(3-2x\right)\\ \Leftrightarrow x=3-2x\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\\ \text{Vậy }S=\left\{1\right\}\)
f) Kiểm tra lại hạng tử thứ 2 ở vế phải.
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)
\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)
=>1/12x=77/57
=>x=308/19
b: =>(x^2-4)(x^2-10)=72
=>x^4-14x^2+40-72=0
=>x^4-14x^2-32=0
=>(x^2-16)(x^2+2)=0
=>x^2-16=0
=>x^2=16
=>x=4 hoặc x=-4
Bài 2:
\(B=\left[\left(\dfrac{x+1-3}{x-2}-3x\right)\cdot\dfrac{x-2}{1-3x}\right]-\dfrac{x^2+4}{x-2}\)
\(=\left(\dfrac{x-2}{x-2}-3x\right)\cdot\dfrac{x-2}{1-3x}-\dfrac{x^2+4}{x-2}\)
\(=x-2-\dfrac{x^2+4}{x-2}=\dfrac{x^2-4x+4-x^2-4}{x-2}=\dfrac{-4x}{x-2}\)
a: \(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\cdot2\)
\(=\dfrac{10}{5}\cdot2=4\)
b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}=1\)
a) \(A=\left(3x-2\right)^2+\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)\)
\(\Leftrightarrow A=\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(\Leftrightarrow A=\left[\left(x+1\right)-\left(3x-2\right)\right]^2\)
\(\Leftrightarrow A=\left(x+1-3x+2\right)^2\)
\(\Leftrightarrow A=\left(3-2x\right)^2\)
Thay \(x=\dfrac{3}{2}\) vào biểu thức A ta được:
\(\left(3-2.\dfrac{3}{2}\right)^2=\left(3-3\right)^2=0^2=0\)
Vậy giá trị của biểu thức A tại \(x=\dfrac{3}{2}\) là 0
b) \(B=\dfrac{x^2y\left(y-x\right)-xy^2\left(x-y\right)}{3y^2-3x^2}\)
\(\Leftrightarrow B=\dfrac{x^2y\left(y-x\right)+xy^2\left(y-x\right)}{3\left(y^2-x^2\right)}\)
\(\Leftrightarrow B=\dfrac{\left(y-x\right)\left(x^2y+xy^2\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(x+y\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(y+x\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy}{3}\)
Thay \(x=-3\) và \(y=\dfrac{1}{2}\) vào biểu thức B ta được:
\(\dfrac{\left(-3\right).\dfrac{1}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{-1}{2}\)
Vậy giá trị của biểu thức B tại \(x=-3\) và \(y=\dfrac{1}{2}\) là \(\dfrac{-1}{2}\)
c) \(C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\) MTC: \(\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)-\left(x-3\right)\left(1-x\right)+2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x^2+3x+x+3\right)-\left(x-x^2-3+3x\right)+\left(2x-2x^2\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{x^2+3x+x+3-x+x^2+3-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2}{x-3}\)
Thay \(x=5\) vào biểu thức C ta được:
\(\dfrac{2}{5-3}=\dfrac{2}{2}=1\)
Vậy giá trị của biểu thức C tại \(x=5\) là 1
a) \(3\left(4x-1\right)-2x\left(5x+2\right)>8x-2\)
\(\Leftrightarrow12x-3-10x^2-4x>8x-2\)
\(\Leftrightarrow-10x^2>5\)
\(\Leftrightarrow x^2< \dfrac{-1}{2}\)(vô lí)
Vậy bất phương trình đã cho vô nghiệm.
h)
\(\dfrac{x+5}{x+7}-1>0\)
\(\Leftrightarrow\dfrac{x+5}{x+7}-\dfrac{x+7}{x+7}>0\)
\(\Leftrightarrow\dfrac{x+5-x-7}{x+7}>0\)
\(\Leftrightarrow\dfrac{-2}{x+7}>0\)
\(\Leftrightarrow x+7< 0\)
\(\Leftrightarrow x< -7\)
g)
\(\dfrac{4-x}{3x+5}\ge0\)
* TH1:
\(4-x\ge0\) và \(3x+5>0\)
\(\Leftrightarrow x\le4\) và \(x>\dfrac{-5}{3}\)
* TH2:
\(4-x\le0\) và \(3x+5< 0\)
\(\Leftrightarrow x\ge4\) và \(x< \dfrac{-5}{3}\) ( loại)
Vậy: \(-\dfrac{5}{3}< x\le4\)
1/ \(P\left(x\right)=x^3-3x^2+5x-2a\)
Để \(P\left(x\right)\) chia hết cho \(x-2\) thì \(P\left(2\right)=0\)
\(\Leftrightarrow8-12+10-2a=0\Leftrightarrow a=3\)
2/Thực hiện phép chia đa thức ta được:
\(x^4-3x^2+ax+b=\left(x^2-3x+4\right)\left(x^2+3x+2\right)+\left(a-6\right)x+b-8\)
Để \(x^4-3x^2+ax+b\) chia hết \(x^2-3x+4\)
\(\Rightarrow\left\{{}\begin{matrix}a-6=0\\b-8=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=6\\b=8\end{matrix}\right.\)
3/ \(\dfrac{a}{x-2}+\dfrac{b}{x+3}=\dfrac{a\left(x+3\right)+b\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(a+b\right)x+3a-2b}{\left(x-2\right)\left(x+3\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\3a-2b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
4/ \(\dfrac{a}{x-1}+\dfrac{b}{\left(x-1\right)^2}=\dfrac{a\left(x-1\right)+b}{\left(x-1\right)^2}=\dfrac{ax+b-a}{\left(x-1\right)^2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b-a=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=8\end{matrix}\right.\)