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,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=> 15x²+5x-6x-2+7x+21-15x²-45x+20=0

<=>39-39x=0

<=>39(1-x)=0

<=>1-x=0

=>x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

=>\(15x^2-6x+5x-2+7x-15^2+21-45x=-20\)

=>\(-39x+19=-20\)

=>\(-39x=-39\)

=>\(x=1\)

vậy x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=>\(15x^2+5x-6x-2+7x-15x^2+21-45x=-20\)

<=>\(-39x+19=0\)

<=>\(-39x=-19\)

<=>\(x=\dfrac{19}{39}\)

Vậy \(x=\dfrac{19}{39}\)

\(b,4x^2-x-5=0\)

\(\Leftrightarrow4x^2-5x+4x-5=0\)

\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)

\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)

Bài 2

\(a,x^3+5x^2+3x-9\)

\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)

\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)

b,\(x^3-7x-6\)

\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)

\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c,\(3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)

\(4x^2-x-5=0\)

<=>  \(4x^2+4x-5x-5=0\)

<=>   \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự lm tiếp

Bài 1:

a)\(28x^3+15x^2+75x+125=0\)

\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)

Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)

\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)

b)\(4x^2-x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)

\(\Rightarrow x=-1;x=\frac{5}{4}\)

Bài 2:

a)\(x^3+5x^2+3x-9\)

\(=\left(x-1\right)\left(x+3\right)^2\)

b)\(x^3-7x-6\)

\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c)\(3x^3-7x^2+17x-5\)

\(=\left(3x-1\right)\left(x^2-2x+5\right)\)

\(4x^2-x-5=0\)

<=> \(4x^2+4x-5x-5=0\)

<=>  \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự giải nốt

\(\left(5x+7\right)⋮\left(3x+2\right)\)

\(3\left(5x+7\right)-5\left(3x+2\right)⋮\left(3x+2\right)\)

\(15x+21-15x-10⋮\left(3x+2\right)\)

\(11⋮\left(3x+2\right)\)

\(\left(3x+2\right)\inƯ\left(11\right)=\left\{1;11\right\}\)

Mà 3x + 2 > 0 nên 3x + 2 = 11

Vậy x = 3

a. \(\left(3x-5\right)^2-\left(x+1\right)^2=0\Leftrightarrow\left(3x-5+x+1\right)\left(3x-5-x-1\right)=0\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\2x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy ...

b. \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

c. \(4x^3-36x=0\Leftrightarrow4x\left(x^2-9\right)=0\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Vậy ...

d. \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-2x+3\right)=0\Leftrightarrow6\left(x-1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy ...

2.(x+5) - x² - 5x = 0

2(x+5) - x(x+5) = 0

(x+5)(2-x) = 0

=> x+5=0 hoặc 2-x=0

=> x=-5 hoặc x=2

\(\frac{x^5-1}{x^3-1}=\frac{ }{x+1}\)