5x-2+35x=(3.4+7).102
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1) Ta có: \(3x\left(2-5x\right)+35x-14=0\)
\(\Leftrightarrow3x\left(2-5x\right)+7\left(5x-2\right)=0\)
\(\Leftrightarrow-3x\left(5x-2\right)+7\left(5x-2\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(-3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\-3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=2\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{5};\dfrac{7}{3}\right\}\)
2) Ta có: \(4x-6+5x\left(3-2x\right)=0\)
\(\Leftrightarrow2\left(2x-3\right)-5x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\5x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{2}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};\dfrac{2}{5}\right\}\)
Lần sau bạn chú ý ghi đầy đủ yêu cầu đề.
Lời giải:
1. $9x^2+4+12x=(3x)^2+2.3x.2+2^2=(3x+2)^2$
2. Đề sai sai. Bạn xem lại
3.
$(16x^2-4xy+y^2)(4x+y)=(4x+y)[(4x)^2-4x.y+y^2]$
$=(4x)^3+y^3=64x^3+y^3$
4.
$(5x-7)(25x^2+35x+49)=(5x-7)[(5x)^2+5x.7+7^2]$
$=(5x)^3-7^3=125x^3-343$
\(9x^2+12x+4=\left(3x+2\right)^2\)
\(\left(16x^2-4xy+y^2\right)\cdot\left(4x+y\right)=64x^3+y^3\)
\(\left(5x-7\right)\left(25x^2+35x+49\right)=125x^3-343\)
\(\left(35x^6y^7-25x^5y^5\right):5x^3y^5-7x^3y^2\)
\(=7x^3y^2-5x^2-7x^3y^2\)
\(=-5x^2\)
\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)
\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)
\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)
Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:
\(E=5x.0+105=105\)
Sửa đề: -103/51*52
\(D=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{1}{51}-\dfrac{1}{52}\)
=1/2-1/52
=26/52-1/52=25/52
b: \(=\dfrac{x^3+6x^2-25}{x\left(x+5\right)\left(x-2\right)}+\dfrac{x+5}{x\left(x-2\right)}\)
\(=\dfrac{x^3+6x^2-25+x^2+10x+25}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x^3+7x^2+10x}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)