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Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
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
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)
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) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
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: =>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
Bài 1:
a) \(\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow x+2=0\) hoặc \(x+4=0\)
\(\Leftrightarrow x=-2\) hoặc \(x=-4\)
b) \(2x^3+\dfrac{3}{2}x^2=0\)
\(\Leftrightarrow x^2\left(2x+\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow x^2=0\) hoặc \(2x+\dfrac{3}{2}=0\)
\(\Leftrightarrow x=0\) hoặc \(x=-\dfrac{3}{4}\)
bài 1
a) (x+2)2-x2+4=0
\(\Leftrightarrow\)x2+4x+4-x2+4=0
\(\Leftrightarrow\)4x+8=0
\(\Leftrightarrow\) 4(x+2)=0
=>x+2=0
\(\Leftrightarrow\)x=-2
vậy x=-2
b) \(2x^3+\dfrac{3}{2}x^2=0\)
\(\Leftrightarrow x^2\left(2x+\dfrac{3}{2}\right)=0\)
=>\(\left[{}\begin{matrix}x^2=0\\2x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\-\dfrac{3}{4}\end{matrix}\right.\)
vậy x=0 hoặc x=-\(\dfrac{3}{4}\)