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\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)
Ta có \(64:59R5\Rightarrow64^n:59R5\)
Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)
Vậy \(A⋮59\)
(\(R\) là dư)
\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
cho hỏi là x=-2 thì x đâu còn \(\ge\) 0 nữa
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy.....
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
=>x-99=0
hay x=99
7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)
=>x+100=0
hay x=-100
8:
Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)
=>200-x=0
hay x=200
\(125-x^6=\left(5\right)^3-\left(x^2\right)^3\)
\(=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)
\(49\left(x-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(x-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left[7x-28\right]^2-\left[3y+6\right]^2\)
\(=\left(7x-28-3y-6\right)\left(7x-28+3y+6\right)\)
\(=\left(7x-3y-34\right)\left(7x-22+3y\right)\)
10) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/86 + 1/85 + 1/84 + 1/83 + 1/4 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy....
a) = (x+3).(x-3)^2-(x-3)(x+3)^2
=(x^2-9)(x-3)-(x^2-9)(x+3)
=(x^2-9)(x-3-x-3)
=-6(x^2-9)
các câu còn lại tương tự
\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+3-\left(x^3-3\right)\)
\(=x^3+3-x^3+3\)
\(=6\)
\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)
\(=x^3-5^3-x^3-5^3\)
\(=-125-125\)
\(=-250\)
Câu hỏi của Nghĩa Nguyễn - Toán lớp 9 - Học toán với OnlineMath