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a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\\ \Leftrightarrow\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}\)
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:
\(\dfrac{2x+3}{5x+2}=\dfrac{1}{2}\Leftrightarrow5x+2=4x+6\Leftrightarrow5x=4x+4\Leftrightarrow x=4\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
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: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
B)=>(3x-1).(5x-34)=(40-5x).(25-3x)
=>15x2-102x-5x+34=1000-120x-125x+15x2
=>15x2-107x+34=1000-245x+15x2
=>15x2-15x2-107x+245x=1000-34
=>0-107x+245x=966
=>138x=966
=>x=7
A,=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(20x^2-16x-34=10x^2+3x-34\)
\(10x^2-19x=0\)
\(x\left(10x-19\right)=0\)
\(\Leftrightarrow x=0\)
hoặc \(10x-19=0\)
\(\Leftrightarrow x=\dfrac{19}{10}\)
Vạy ..............
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow x+5=42\)
\(\Leftrightarrow x=37\)
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