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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 \(x\le-\dfrac{8}{3}\)
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
Được câu d rồi mà không biết có đúng không.
\(d)\left(x+3\right)\left(x+2\right)+3x=4\left(x+\dfrac{3}{4}\right)\)
\(x^2+2x+3x+6+3x=4x+3\)
\(x^2+8x+6-4x=3\)
\(x^2+4x+4-4+6=3\)
\(\left(x+2\right)^2+2=3\)
\(\left(x+2\right)^2=3-2=1^2\)
Ta có:
Vậy: \(x=-1\) hoặc \(x=-3\)
\(a)\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)
\(3x^3-4x^2+2x-1-3x^3+4x^2\) \(=\dfrac{5}{2}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{5}{2}+1\) \(=\dfrac{7}{2}\)
\(x=\dfrac{7}{2}:2=\dfrac{7}{2}\cdot\dfrac{1}{2}\)
Vậy: \(x=\dfrac{7}{4}\)
\(b)2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(2x^2+3x^2-3=5x^2+5x\)
\(5x^2-3=5x^2+5x\)
\(-3=5x^2-5x^2+5x\)
\(-5x=3\)
Vậy: \(x=-\dfrac{3}{5}\)
\(c)\left(x^2-x+1\right)\left(x+1\right)-\left(x^3-3x\right)=15\)
\(x^3+1-x^3+3x=15\)
\(3x+1=15\)
\(3x=15-1=14\)
Vậy: \(x=\dfrac{14}{3}\)
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, \(xy\left(x+y\right)-x^2\left(x+y\right)-y^2\left(x-y\right)\)
\(=x^2y+xy^2-x^3-x^2y-xy^2+y^3\)
\(=y^3-x^3\)
b, \(x^2-x^2\left(5x+1\right)+x\left(x-3\right)\)
\(=x^2-5x^3-x^2+x^2-3x\)
\(=-5x^3+x^2-3x\)
Chúc bạn học tốt!!!
c, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2-5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=-10x^2-11x+24\)
d, \(\dfrac{1}{2}\left(x+4\right)+\dfrac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2}x+2+3x^3-\dfrac{3}{2}x^2-x^3-\dfrac{1}{2}x\)
\(=-x^3+\left(3x^2-\dfrac{3}{2}x^2\right)+\left(\dfrac{1}{2}x-\dfrac{1}{2}x\right)+2\)
\(=-x^3+\dfrac{3}{2}x^2+2\)
\(=-\left(x^3-\dfrac{3}{2}x^2-2\right)=-\left(x^3-2x^2+\dfrac{1}{2}x^2-x+x-2\right)\)
\(=-\left[\left(x^3-2x^2\right)+\left(\dfrac{1}{2}x^2-x\right)+\left(x-2\right)\right]\)
\(=-\left[x^2.\left(x-2\right)+\dfrac{1}{2}x.\left(x-2\right)+\left(x-2\right)\right]\)
\(=-\left[\left(x-2\right).\left(x^2+\dfrac{1}{2}x+1\right)\right]\)
Chúc bạn học tốt!!!
1. a. \(\left(a+b\right)^2-4\)
\(=\left(a+b+2\right)\left(a+b-2\right)\)
b. \(4a^2+8ab-3a-6b\)
\(=4a\left(a+b\right)-3\left(a+b\right)\)
\(=\left(4a-3\right)\left(a+b\right)\)
c. \(a^2+b^2-c^2-2ab\)
\(=\left(a+b\right)^2-c^2\)
\(=\left(a+b+c\right)\left(a+b-c\right)\)
d. \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(5x-3\right)\left(x-y\right)\)
2. a. \(\dfrac{1-x}{x}+\dfrac{x}{1+x}\)
\(=\dfrac{1-x^2}{x\left(1+x\right)}+\dfrac{x^2}{x\left(1+x\right)}\)
\(=\dfrac{1-x^2+x^2}{x\left(1+x\right)}=\dfrac{1}{x\left(1+x\right)}\)
b. \(\dfrac{4}{x+2}+\dfrac{3}{2-x}+\dfrac{12}{x^2-4}\)
\(=\dfrac{4x-8}{\left(x+2\right)\left(x-2\right)}-\dfrac{3x+6}{\left(x+2\right)\left(x-2\right)}+\dfrac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{4x-8-3x-6+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)
3. \(\dfrac{x}{3x+y}-\dfrac{x}{3x-y}-\dfrac{2x^2}{xy^2-9x^3}\)
\(=\dfrac{3x^3-x^2y}{x\left(3x+y\right)\left(3x-y\right)}-\dfrac{3x^3+x^2y}{x\left(3x+y\right)\left(3x-y\right)}-\dfrac{2x^2}{x\left(y-3x\right)\left(y+3x\right)}\)
\(=\dfrac{3x^3-x^2y-3x^3-x^2y+2x^2}{x\left(3x+y\right)\left(3x-y\right)}\)
\(=\dfrac{-x^2y+2x^2}{x\left(3x+y\right)\left(3x-y\right)}\)
\(=\dfrac{-xy+2x}{\left(3x+y\right)\left(3x-y\right)}\)
Thay x = 1 và y = 2 vào phân thức ta được:
\(=-\dfrac{2+2.2}{\left(3+2\right)\left(3-2\right)}=-\dfrac{6}{5}\)
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
Bài 2:
\(=\dfrac{x^2\left(x^2+4\right)-2x\left(x^2+4\right)}{x^2+4}=x^2-2x\)
Bài 1:
a: \(=\left(\dfrac{2}{3}:\dfrac{-1}{9}\right)\cdot x^4y^2z^6=-6x^4y^2z^6\)
b: \(=-12x^8-21x^5\)
c: =x^3+8
d: \(=125x^3-75x^2+15x-1\)
a) (3x - 2)2 - (1 + 5x)2
= (3x - 2 - 1 - 5x)(3x - 2 + 1 + 5x)
= (-2x - 3)(8x - 1)
b) (3x + 4)(3x - 4) - (5 - x)2
= (3x)2 - 42 - (25 - 10x + x2)
= 9x2 - 16 - 25 + 10x - x2
= 8x2 + 10x - 41
c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)
\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)
\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)
\(=4x+25\)
a: =9x^2-12x+4-25x^2-10x-1
=-16x^2-22x+3
b: =9x^2-16-x^2+10x-25
=8x^2+10x-41
c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)