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a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}=\frac{6.2}{2x\left(x+4\right)}+\frac{3x}{2x\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(x+4\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)
c) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5.y}{2y\left(y+2\right)}+\frac{2\left(y-2\right)}{2y\left(y+2\right)}=\frac{-5y+2y-4}{2y\left(y+2\right)}=\frac{-3y-4}{2y\left(y+2\right)}\)
d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x\left(x-2y\right)}-\frac{3}{x\left(x-2y\right)}=\frac{x-1-3}{x\left(x-2y\right)}=\frac{x-4}{x\left(x-2y\right)}\)
\(a,x^3-x^2-12x+45=0\)
\(\left(x-3\right)\left(x-3\right)\left(x+5\right)=0\)
\(x=3;3;-5\)
\(b,2x^3-5x^2+8x-5=0\)
\(\left(2x^2-3x+5\right)\left(x-1\right)=0\)
\(x=1\)
lm 1 câu đã chán ngắt , giải mấy câu nữa não tớ nổ bùmmm , tớ bt đây là trang web để hc nhưng tạo nên tiếng cười là chính nha ^^
HIHI, bài này thì bó tay lẫn cả chân
Vì mới học xong lớp 6 hoi.
Học tốt nha, nếu ko ai giải thì thử vào câu hỏi tương tự thử
Nha, học tốt !
#)Giải:
-Không sao mình biết cách làm mà, mình chỉ thử lòng ae thui !
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0
Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)
A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)
A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)
b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)
B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{x+3y}{x\left(x-3y\right)}\)
\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)
\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)
\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)
\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)
Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(\left(4+2x\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}4+2x=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=-4\\x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
vậy ta chọn : B
1.
a. $A=\frac{x^3-x+2}{x-2}=\frac{x^2(x-2)+2x(x-2)+4(x-2)+10}{x-2}$
$=x^2+2x+4+\frac{10}{x-2}$
Với $x$ nguyên, để $A$ nguyên thì $\frac{10}{x-2}$ là số nguyên.
Khi $x$ nguyên, điều này xảy ra khi $10\vdots x-2$
$\Rightarrow x-2\in \left\{\pm 1; \pm 2; \pm 5; \pm 10\right\}$
$\Rightarrow x\in \left\{3; 1; 4; 0; 7; -3; 12; -8\right\}$
b.
\(B=\frac{2x^2+5x+8}{2x+1}=\frac{x(2x+1)+3x+8}{2x+1}=x+\frac{3x+8}{2x+1}\)
Với $x$ nguyên, để $B$ nguyên thì $3x+8\vdots 2x+1$
$\Rightarrow 2(3x+8)\vdots 2x+1$
$\Rightarrow 3(2x+1)+13\vdots 2x+1$
$\Rightarrow 13\vdots 2x+1$
$\Rightarrow 2x+1\in \left\{\pm 1; \pm 13\right\}$
$\Rightarrow x\in \left\{0; -1; 6; -7\right\}$
Bài 2:
$P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}=\frac{(2x-1)^3}{(2x-1)^2}=2x-1$
Với $x$ nguyên thì $2x-1$ cũng là số nguyên.
$\Rightarrow P$ nguyên với mọi $x$ nguyên.