Cho phân số: \(\dfrac{a}{b}\) và \(\dfrac{a-x}{b-x}\)=\(\dfrac{a}{b}\)
Chứng minh \(\dfrac{-x}{y}\)=\(\dfrac{a}{b}\)
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a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)
\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)
\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)
\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)
\(=\dfrac{3a-1}{2}\)
\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)
b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)
\(=\dfrac{1}{b-3}\)
\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)
\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)
\(=>\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\)
\(=>\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}=>\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}=\dfrac{5}{3}\)
vậy \(\dfrac{a}{b}=\dfrac{5}{3}\)
\(\Rightarrow\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{3}\\ \Rightarrow a=5;b=3\)
a: Ta có: \(A=\dfrac{1}{2}\)
\(\Leftrightarrow x+2=2x-6\)
\(\Leftrightarrow-x=-8\)
hay x=8
Thay x=8 vào B,ta được:
\(B=-\dfrac{2}{8+2}=-\dfrac{2}{10}=-\dfrac{1}{5}\)
\(\dfrac{a-x}{b-y}=\dfrac{a}{b}\)
\(\Rightarrow\dfrac{a-x}{a}=\dfrac{b-y}{b}\)
\(\Rightarrow1-\dfrac{x}{a}=1-\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{a}{b}\)
a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\)
b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\)
\(\dfrac{x}{3}=\dfrac{-14}{15}\)
\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)
\(x=\dfrac{-2}{7}+\dfrac{9}{7}\)
\(x=1\)
\(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\left(1\right)\)
a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
\(\left(1\right)\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x+1}\)
b) Để \(A=-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x^2}{x+1}=-\dfrac{1}{2}\left(x\ne-1\right)\)
\(\Leftrightarrow2x^2=-\left(x+1\right)\)
\(\Leftrightarrow2x^2+x+1=0\)
\(\Delta=1-8=-7< 0\)
Nên phương trình trên vô nghiệm \(\left(x\in\varnothing\right)\)
c) Để \(A< 1\)
\(\Leftrightarrow\dfrac{x^2}{x+1}< 1\)
\(\Leftrightarrow x^2< x+1\left(x\ne-1\right)\)
\(\Leftrightarrow x^2-x-1< 0\)
\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1< 0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}< 0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2< \dfrac{5}{4}\)
\(\Leftrightarrow-\dfrac{\sqrt[]{5}}{2}< x-\dfrac{1}{2}< \dfrac{\sqrt[]{5}}{2}\)
\(\Leftrightarrow\dfrac{-\sqrt[]{5}+1}{2}< x< \dfrac{\sqrt[]{5}+1}{2}\)
d) Để A nguyên
\(\Leftrightarrow\dfrac{x^2}{x+1}\in Z\)
\(\Leftrightarrow x^2⋮x+1\)
\(\Leftrightarrow x^2-x\left(x+1\right)⋮x+1\)
\(\Leftrightarrow x^2-x^2+x⋮x+1\)
\(\Leftrightarrow x⋮x+1\)
\(\Leftrightarrow x-x-1⋮x+1\)
\(\Leftrightarrow-1⋮x+1\)
\(\Leftrightarrow x+1\in\left\{-1;1\right\}\)
\(\Leftrightarrow x\in\left\{-2;0\right\}\left(x\in Z\right)\)
\(VT=\dfrac{a^2}{b+ab^2c}+\dfrac{b^2}{b+abc^2}+\dfrac{c^2}{c+a^2bc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}=\dfrac{9}{3+3abc}\)
\(VT\ge\dfrac{9}{3+\dfrac{\left(a+b+c\right)^3}{9}}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ngô Hải Nam ơi bn trả lời giúp mik ik
bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên
Lời giải:
$\frac{a-x}{b-y}=\frac{a}{b}$
$\Rightarrow b(a-x)=a(b-y)$
$\Rightarrow ab-bx=ab-ay$
$\Rightarrow bx=ay$
$\Rightarrow \frac{a}{b}=\frac{x}{y}$