-15+19x=9+7x
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https://olm.vn/hoi-dap/tim-kiem?q=%7C2x-9%7C+%7C3x-13%7C-%7C14-5x%7C+%7C19-(4x+12)%7C=56t%C3%ACm+x&id=191953
\(\Leftrightarrow\left[{}\begin{matrix}14-19x=7x-15\\19x-14=7x-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}26x=29\\12x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{26}\\x=-\dfrac{1}{12}\end{matrix}\right.\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\frac{\left(x-\frac{2}{5}\right)\left(x+3\right)}{\left(x+\frac{1}{3}\right)\left(x+3\right)}\)
\(=\frac{x-\frac{2}{5}}{x+\frac{1}{3}}\)
=\(\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
=\(\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
=\(\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
=\(\frac{2x^2-6x+5x-15}{3x^2-9x-x+3}\)
=\(\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
=\(\frac{2x+5}{3x-1}\)
2x3 - 7x2 - 12x + 45 = 2x3 - 6x2 - x2 + 3x - 15x + 45
= 2x2(x - 3) - x(x - 3) - 15(x - 3)
= (x - 3)(2x2 - x - 15)
= (x - 3)(2x2 - 6x + 5x - 15)
= (x - 3)((2x(x - 3) + 5(x - 3))
= (x - 3)2(2x + 5)
3x3 - 19x2 +33x - 9 = 3x3 -9x2 -10x2 + 30x +3x - 9
= 3x2(x - 3) - 10x(x - 3) + 3(x - 3)
= (x - 3)(3x2 - 10x + 3)
= (x - 3)(3x2 -9x - x +3)
= (x - 3)((3x(x-3) - (x - 3))
=(x - 3)2(3x - 1)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)
Ta có tử bằng:2x3-7x2-12x+45
=(2x3-6x2)-(x2-3x)-(15x-45)
=2x2(x-3)-x(x-3)-15(x-3)
=(x-3)(2x2-x-15)
=(x-3)(2x2-6x+5x-15)
=(x-3)2(2x+5) (1)
Ta có mẫu bằng:3x3-19x2+33x-9
=(3x3-x2)-(19x2-6x)+(27x-9)
=x2(3x-1)-6x(3x-1)+9(3x-1)
=(3x-1)(x2-6x+9)
=(3x-1)(x-3)2 (2)
Thay (1) và (2) vào phân thức ,ta có:
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}=\frac{2x+5}{3x-1}\)
Ta có: \(C=\frac{3x^2-7x^2-12+45}{3x^3-19x^2+33x-9}\) ĐKXĐ: x khác 3, 1/3
\(=\frac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}\)
\(=\frac{2x+5}{3x-1}\)
Để C>0, ta có:
-5/2<x<1/3 (thỏa mãn ĐKXĐ)
Lời giải:
Ta có:
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\text{TS}}{\text{MS}}\)
Xét \(\text{TS}=2x^2(x-3)-x(x-3)-15(x-3)\)
\(=(x-3)(2x^2-x-15)=(x-3)[2x(x-3)+5(x-3)]\)
\(=(x-3)(x-3)(2x+5)=(x-3)^2(2x+5)\)
Xét \(\text{MS}=3x^2(x-3)-10x(x-3)+3(x-3)\)
\(=(x-3)(3x^2-10x+3)=(x-3)[3x(x-3)-(x-3)]\)
\(=(x-3)(x-3)(3x-1)=(x-3)^2(3x-1)\)
Do đó:
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(x-3)^2(3x-1)}=\frac{2x+5}{3x-1}\)
Xét tử thức ta có
2x3-7x2-12x+45
= 2x3+5x2-12x2-30x+18x+45
= x2(2x+5)-6x(2x+5)+9(2x+5)
= (2x+5)(x2-6x+9)
= (2x+5)(x-3)2 (1)
Xét mẫu thức ta có
3x3-19x2+33x-9
= 3x3-x2-18x2+6x+27x-9
= x2(3x-1)-6x(3x-1)+9(3x-1)
= (3x-1)(x2-6x+9)
= (3x-1)(x-3)2 (2)
Thay (1) và (2) vào A ta được\(A=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)
a, Để phân thức trên có nghĩa thì:
\(3x^3-19x^2+33x-9\ne0\)
\(\Rightarrow3x^3-9x^2-10x^2+30x+3x-9\ne0\)
\(\Rightarrow3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)\ne0\)
\(\Rightarrow\left(x-3\right)\left(3x^2-10x+3\right)\ne0\)
\(\Rightarrow\left(x-3\right).\left[3x^2-9x-x+3\right]\ne0\)
\(\Rightarrow\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]\ne0\)
\(\Rightarrow\left(x-3\right)^2.\left(3x-1\right)\ne0\)
\(\Rightarrow\hept{\begin{cases}x-3\ne0\\3x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne3\\x\ne\frac{1}{3}\end{cases}}}\)
1. \(x^2+2x-15=0\)
\(\Rightarrow x^2+2x+1^2-16=0\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\).
2. \(x^2-7x-44=0\)
\(\Rightarrow x^2-2.x.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{49}{4}-44=0\)
\(\Rightarrow\left(x-\dfrac{7}{4}\right)^2=\left(\dfrac{15}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{7}{4}=\dfrac{15}{2}\\x-\dfrac{7}{4}=-\dfrac{15}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{37}{4}\\x=\dfrac{-23}{4}\end{matrix}\right.\).
3.4 Tương tự.
2) hãy dành 5(s)
\(x^2-7x-44=0\Rightarrow\left(x^2+4x\right)-\left(11x+44\right)=0\)
\(x\left(x+4\right)-11\left(x+4\right)=0\)
\(\left(x+4\right)\left(x-11\right)=0\)\(\left[{}\begin{matrix}x=-4\\x=11\end{matrix}\right.\)
`-15 + 19x = 9 + 7x`
`=> 12x = 24`
`=> x = 2`.