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a) \(\left(3x-5\right)\left(9x^2+15x+25\right)\)
\(=\left(3x\right)^3-5^3\)
\(=27x^3-125\)
b) \(\left(2x+7\right)\left(x^2-14x+49\right)-2x\left(2x-1\right)\left(2x+1\right)\)
\(=2x^3-28x^2+98x+7x^2-98x+343-2x\left(4x^2-1\right)\)
\(=2x^3-28x^2+7x^2+343-8x^3+2x\)
\(=-6x^3-21x^2+343+2x\)
c) \(\left(4x-7\right)\left(16x^2+28x+49\right)\left(3x+1\right)\left(9x^2-3x+1\right)-9x\left(3x^2-1\right)\)
\(=\left(64x^3-343\right)\left(3x+1\right)\left(9x^2-3x+1\right)-27x^3+9x\)
\(=\left(6x^3-343\right)\left(27x^3+1\right)-27x^3+9x\)
\(=1728x^6+64x^3-9261x^3-343-27x^3+9x\)
\(=1728x^6-9224x^3-343+9x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này chắc giải phương trình . Lần sau nếu bn muôn hỏi bài nào thì ghi rõ tên đề bài nhé, chứ như thế này ko biết đề bài như thế nào đâu .Đây mik làm đại nhé.
\(\frac{3x^5\left(4x^2+5\right)^2}{\left(4x^2+5\right)^2}-\frac{x\left(3x^4+7\right)^2}{3x^4+7}=2x-5\)
\(\Leftrightarrow3x^5-x\left(3x^4+7\right)=2x-5\)
\(\Leftrightarrow3x^5-3x^5-7x=2x-5\)
\(\Leftrightarrow-7x-2x=-5\)
\(\Leftrightarrow-9x=-5\Leftrightarrow x=\frac{5}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+5\right)\left(5x+5\right)}-\dfrac{x^2-25}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(3x-3-x\right)\left(3x-3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{x^2-25}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{3\left(x+5\right)\cdot5\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(x+5\right)}{5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2-\left(x+5\right)\left(x+1\right)}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-6x-5}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{3x^2+18x+31}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{3\left(x+3\right)\left(3x^2+18x+31\right)-\left(2x-3\right)\left(4x-3\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{\left(3x+9\right)\left(3x^2+18x+31\right)-\left(8x^2-18x+9\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3+8x^2-18x^2-18x+9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3-10x^2-9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{x^3+91x^2+264x+270}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt
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a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
Rút gọn biểu thức:
\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)
a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
Vậy: \(S=\left\{3;20\right\}\)
c) Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
a: =>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: =>(x-3)(x+20)=0
=>x=3 hoặc x=-20
c: =>4x+2=0
hay x=-1/2
d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0
=>x=-7/2 hoặc x=5 hoặc x=-1/5
Ta có : ( 2x - 5).(3x+7 ) = 4x2 -25
<=>6x2 + 14x - 15x -35 = 4x2 -25
<=> 2x2 -x -
Mình xin lỗi bạn nha ,do bất cẩn quá nên mình lỡ nhấn gửi câu trả lời ,để mình làm lại nhé !!!
Ta có :( 2x -5 )(3x + 7)=4x2 -25
<=> 6x2 + 14 - 15x-35 =4x2 -25
<=> 2x2 -x - 10 = 0
<=> ( 2x-5)(x+2)=0
<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-2\end{cases}}\)
Vậy x = \(\left\{\frac{5}{2};-2\right\}\)