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Bài 1:
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(2x^2+5x+3=2x^2+2x+3x+3=2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(2x+3\right)\)
c) \(x^2-10x+16=x^2-2x-8x+16=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)
d) \(4x^2+9x+5=4x^2+4x+5x+5=4x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(4x+5\right)\)
Bài 2:
không rõ đề --> k lm
a) \(\left|2x-3\right|-\dfrac{5}{2}=\dfrac{1}{3}\)
\(\left|2x-3\right|=\dfrac{1}{3}+\dfrac{5}{2}=\dfrac{2}{6}+\dfrac{15}{6}\)
\(\left|2x-3\right|=\dfrac{17}{6}\)
\(+)2x-3=\dfrac{17}{6}\Rightarrow2x=\dfrac{35}{6}\Rightarrow x=\dfrac{35}{12}\)
\(+)2x-3=\dfrac{-17}{6}\Rightarrow2x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{12}\)
vậy...
\(\left|x-1\right|+3x=1\\ \Rightarrow\left|x-1\right|=1-3x\\ \Rightarrow\left\{{}\begin{matrix}x-1=1-3x\\x-1=-1+3x\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x=2\\-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
Dấu ngoặc vuông nhé
thánh bấm nhầm
để \(\dfrac{x+1}{x-1}\)nguyên thì
(x+1)⋮(x-1)
=> (x+1)-(x-1)⋮(x-1)
=> (x+1-x+1)⋮(x-1)
=> 2⋮(x+1)
=> x+1 ∈Ư (2)={-2;-1;1;2}
ta có bảng sau
| x+1 | -2 | -1 | 1 | 2 |
| x | -3 | -2 | 0 |
1 |
vậy để \(\dfrac{x+1}{x-1}\)thì x ∈{-3;-2;0;1}
\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\\ \Leftrightarrow\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}\)
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)
=>\(4x+6=\dfrac{1}{2}\left(10x+4\right)\)
=>4x+6=5x+2
=>x=6-2=4
Vậy x=4
Mình sẽ làm cách dãy tỉ số bằng nhau ,vì nhân sẽ khá là rối =.=
\(\dfrac{2x+3}{5x+2}=\dfrac{2\left(2x+3\right)}{2\left(5x+2\right)}=\dfrac{4x+6}{10x+4}\)
Hay \(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)
Thay vào ta có:
\(\dfrac{2x+3}{5x+2}=\dfrac{1}{2}\Leftrightarrow5x+2=4x+6\Leftrightarrow5x=4x+4\Leftrightarrow x=4\)
b: \(ab\cdot bc\cdot ac=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow\left(abc\right)^2=\dfrac{1}{4}\)
Trường hợp 1: abc=1/2
\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{1}{2}:\dfrac{1}{2}=1\\a=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{3}{4}\\b=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)
Trường hợp 2: abc=-1/2
\(\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a=-\dfrac{3}{4}\\b=-\dfrac{2}{3}\end{matrix}\right.\)
c: Theo đề, ta có: \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{1}\\\dfrac{y-2}{3}=\dfrac{z-3}{4}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot6-3\cdot6+3\cdot4}=\dfrac{45}{6}=\dfrac{15}{2}\)
Do đó: x-1=45; y-2=45/2; z-3=30
=>x=46; y=49/2; z=33
a: =>13/15x=3/4-1/2=1/4
=>x=15/52
b: =>x-3=4
=>x=7
c: =>2x+1=9
=>2x=8
=>x=4
d: =>x+3=-2
=>x=-5
e: =>(x+6)(x-4)=0
=>x=4 hoặc x=-6
f: =>(x-3)(x-7)=0
=>x=3 hoặc x=7
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
Ta có :
\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\)
\(\Leftrightarrow2x\left(10x+2\right)+3\left(10x+2\right)=4x\left(5x+2\right)+5\left(5x+2\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+8x+25x+10\)
\(\Leftrightarrow20x^2+34x+6=20x^2+33x+10\)
\(\Leftrightarrow20x^2+34x+6-20x^2-33x-10=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
Vậy x = 4