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CÂU 1:
a) \(2x+4+x^2=-2x+x-3x+2x\)
\(\Leftrightarrow2x+4+x^2=-2x\)
\(\Leftrightarrow x^2+4x+4=0\)
\(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
b) \(2x^2-5x-x=x^2+6x\)
\(\Leftrightarrow2x^2-5x-x-x^2-6x=0\)
\(\Leftrightarrow3x^2-12x=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
Hoặc \(3x=0\Leftrightarrow x=0\)
Hoặc \(x-4=0\Leftrightarrow x=4\)
\(a^3+b^3=2.\left(c^3-8d^3\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3c^2-15d^3⋮3\)
\(a^3+b^3+c^3+d^3-\left(a+b+c+d\right)⋮3\Rightarrow a+b+c+d⋮3\)
tự c/n \(a^3+b^3+c^3+d^3-\left(a+b+c+d\right)⋮3\)nha, gợi ý 1 cái rồi còn lại tương tự
\(a^3-a=a.\left(a^2-1\right)=a.\left(a-1\right).\left(a+1\right)\)chia hết cho 3( vì a,b,c,d thuộc Z)
ợ mk ngu toán lắm, bn lm ơn giải rõ ràng ra hộ nhaaa
a) \(ĐKXĐ:x\ne\pm3\)
\(A=\frac{5}{x+3}-\frac{2}{3-x}+\frac{3x^2-2x-9}{x^2-9}\)
\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x}{x+3}\)
b) Khi \(\left|x-2\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\2-x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Thay x = 1 vào A, ta được :
\(A=\frac{-3}{1+3}=\frac{-3}{4}\)
Vậy khi \(\left|x-2\right|=1\Leftrightarrow A=-\frac{3}{4}\)
c) Để \(A\inℤ\)
\(\Leftrightarrow\frac{-3x}{x+3}\inℤ\)
\(\Leftrightarrow-3x⋮x+3\)
\(\Leftrightarrow-3\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow9⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)
Bài 1:
a)
\(M=\frac{2x}{x+3}+\frac{x+1}{x-3}+\frac{11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{2x\cdot\left(x-3\right)}{x^2-9}+\frac{\left(x+1\right)\cdot\left(x+3\right)}{x^2-9}+\frac{11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{2x^2-6x}{x^2-9}+\frac{x^2+4x+3}{x^2-9}+\frac{11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{2x^2-6x+x^2+4x+3+11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{3x^2+9x}{x^2-9}\\ \Leftrightarrow M=\frac{3x\cdot\left(x+3\right)}{\left(x+3\right)\cdot\left(x-3\right)}\\ \Rightarrow M=\frac{3x}{x-3}\)
b) Theo đề bài có:
\(M=\frac{3}{2}\\ \Leftrightarrow M=\frac{3}{x-3}=\frac{3}{2}\\ \Rightarrow x-3=\frac{3\cdot2}{3}=2\\ \Rightarrow x=2+3=5\\ \Rightarrow x=5\)
Vậy x = 5 thì \(M=\frac{3}{2}\)
c) Theo đề bài có:
\(\frac{1}{M}\le\frac{1}{6}\\ \Leftrightarrow\frac{1}{\frac{3x}{x-3}}=\frac{1}{3x^2-9x}\le\frac{1}{6}\\ \Rightarrow3x^2-9x\le\frac{6\cdot1}{1}\le6\\ \Leftrightarrow x^2-3x\le2\\ \Leftrightarrow\left(x-1.5\right)\cdot\left(x-15\right)\le4.25\\ \Rightarrow x-1.5\le\frac{\sqrt{17}}{2}\\ \Rightarrow x\le\frac{\sqrt{17}}{2}+1.5\le\frac{3+\sqrt{17}}{2}\)
Vậy \(x\le\frac{3+\sqrt{17}}{2}\) thì \(\frac{1}{M}\le\frac{1}{6}\)