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Bài V:
-ĐKXĐ: \(x\ne\pm1\).
\(\dfrac{m}{x-1}+\dfrac{x}{x+1}=\dfrac{x^2}{x^2-1}\)
\(\Leftrightarrow\dfrac{m\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow mx+m+x^2-x=x^2\)
\(\Leftrightarrow m\left(x+1\right)=x\)
\(\Leftrightarrow m=\dfrac{x}{x+1}\)
-Vì m,x nguyên:
\(\Rightarrow x⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1-1\right)⋮\left(x+1\right)\)
\(\Rightarrow-1⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in\left\{1;-1\right\}\)
\(\Rightarrow x\in\left\{0;-2\right\}\) (nhận)
*\(x=0\Rightarrow m=\dfrac{x}{x+1}=\dfrac{0}{0+1}=0\)
\(x=-2\Rightarrow m=\dfrac{x}{x+1}=\dfrac{-2}{-2+1}=1\)
-Vậy với \(m=0\) thì \(S=\left\{0\right\}\)
với \(m=1\) thì \(S=\left\{-2\right\}\)
\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
Bài cuối mình không thấy rõ đề nhưng mình đoán là thế này bạn nhé.
Câu 3:
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)
\(=2x^2+14x-3x^2-3x\)
\(=-x^2+11x\)
Câu 2:
a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)
\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)
\(=-2x^3+3x-4\)
b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)
\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)
\(=2x^2y^2-3x+\dfrac{3}{2}y\)
c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)
\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)
\(=x^2-8x+3\)
d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)
\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)
\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)
\(x^4-4x^2+x^2-4x=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)+x\left(x-4\right)=0\)
\(\Leftrightarrow x\left(x^3-4x+x-4\right)=0\)
\(\Leftrightarrow x\left(x^3-3x-4\right)=0\)
hay x=0
\(\Leftrightarrow x\left(x^3-3x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^3-3x-4=0\end{matrix}\right.\\ \Leftrightarrow x=0\)
\(\left(-2+x^2\right)\left(-2+x^2\right)\left(-2+x^2\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
\(\Leftrightarrow\left(-2+x^2\right)^5=1\)
\(\Leftrightarrow-2+x^2=1\)
\(\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)
1: \(\dfrac{4x^3-2x^2-3x+1}{x-2}\)
\(=\dfrac{4x^3-8x^2+6x^2-12x+9x-18+19}{x-2}\)
\(=4x^2+6x+9+\dfrac{19}{x-2}\)
2: \(\dfrac{2x^4-x^3-3x^2-2x}{x-2}\)
\(=\dfrac{2x^4-4x^3+5x^3-10x^2+7x^2-14x+12x-24+24}{x-2}\)
\(=2x^3+5x^2+7x+12+\dfrac{24}{x-2}\)
a, ta có A(x)=2x3+7x2+ax+b
=(2x3+2x2+2x)+(5x2+5x+5)+ax-7x+b-5
=2x(x2+x+1)+5(x2+x+1)+(a-7)x+(b-5)
=(x2+x+1)(2x+5)+(a-7)x+(b-5)
ta có: (x2+x+1)(2x+5)⋮B(x)
→để A(x)⋮B(x) thì (a-7)x+(b-5)=0
→\(\left\{{}\begin{matrix}a-7=0\\b-5=0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}a=7\\b=5\end{matrix}\right.\)
vậy ....
mk trình bày hơi tắt xíu
bn cố gắng dịch nhé
cả phần c bài 3 ạ
Bài 4 :
a, \(D=\left(\frac{2}{x-3}+\frac{1}{x+3}\right):\frac{x+1}{x-3}\)ĐK : \(x\ne\pm3;-1\)
\(=\left(\frac{2x+6+x-3}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+1}{x-3}=\frac{3\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+1\right)}=\frac{3}{x+3}\)
b, \(D=\frac{3}{x+3}=\frac{x}{6}\Rightarrow x^2+3x=18\Leftrightarrow\left(x-3\right)\left(x+6\right)=0\Leftrightarrow x=3\left(ktm\right);x=-6\left(tm\right)\)
c, \(D=\frac{3}{x+3}< \frac{x}{x+3}\Leftrightarrow\frac{3-x}{x+3}< 0\Leftrightarrow\frac{x-3}{x+3}>0\)
TH1 : \(\hept{\begin{cases}x-3>0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>3\\x>-3\end{cases}\Leftrightarrow x>3}}\)
TH2 : \(\hept{\begin{cases}x-3< 0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x< -3\end{cases}}\Leftrightarrow x< -3}\)
Vậy x > 3 ; x < -3
d, Để \(\frac{3}{x+3}\in Z\Rightarrow x+3\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
3c làm tương tự 4d em nhé