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19 tháng 8 2019

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

19 tháng 8 2019

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

19 tháng 8 2019

1) \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\left(\frac{a}{2}+b\right)^2+\left(\frac{a}{2}-b\right)^2\)

\(=\left(\frac{a}{2}+b\right)\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{b}b+b^2\right]+\left(\frac{a}{2}-b\right)\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a}{2}\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+b\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+\frac{a}{2}\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)\(-b\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a^3}{8}+\frac{a^2b}{2}+\frac{ab^2}{2}+\frac{ba^2}{4}+b^2a+b^3+\frac{a^3}{8}-\frac{a^2b}{2}+\frac{ab^2}{2}-\frac{ba^2}{4}+b^2a-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

2) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3x^2.1+3.x.1^2-1^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

3) \(A=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(A=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9\)

\(A=8\)

Vậy: biểu thức không phụ thuộc vào biến

19 tháng 8 2019

1) \(\left(x+5\right)^3-x^3-125\)

\(=\left(x+5\right)\left(x^2+2x.5+5^2\right)-x^3-125\)

\(=x\left(x^2+2x.5+5^2\right)+5\left(x^2+2x.5+5^2\right)-x^3-125\)

\(=x^3+10x^2+25x+5x^2+50x+125-x^3-125\)

\(=15x^2+75x\)

2) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-4x^2+4x-2x^2+8x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow24x+10=0\)

\(\Leftrightarrow24x=0-10\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\frac{10}{24}=-\frac{5}{12}\)

\(\Rightarrow x=-\frac{5}{12}\)

3) \(\left(x-1\right)^3-x^3+3x^2-3x+1\)

\(=\left(x-1\right)\left(x^2-2x+1\right)-x^3+3x^2-3x+1\)

\(=x\left(x^2-2x+1\right)-\left(x^2-2x+1\right)-x^3+3x^2-3x+1\)

\(=x^3-2x^2+x-x^2+2x-1-x^3-3x^2-3x+1\)

\(=0\)

Vậy: biểu thức không phụ thuộc vào biến

25 tháng 2 2020

c/C=\(\frac{2x^2+2x}{1-x}-\frac{x}{x-1}=\frac{2x^2+2x+x}{1-x}=\frac{2x^2+3x}{1-x}\)

d/C thuộc Z thì C=\(\frac{\left(2x^2-2x\right)+\left(5x-5\right)+5}{1-x}=\frac{-2x\left(1-x\right)-5\left(1-x\right)+5}{1-x}=-2x-5+\frac{5}{1-x}\Rightarrow1-x\in\left(+-1,+-5\right)\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-4\\x=6\end{matrix}\right.\)

25 tháng 2 2020

a/A đã rút gọn B=\(\frac{1-2x}{x^2-3x+2}+\frac{x+1}{x-2}=\frac{1-2x}{\left(x-1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\frac{1-2x+x^2-1}{\left(x-1\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{x}{x-1}\)b/\(\left|x-2\right|=3\Rightarrow\left\{{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}B=\frac{2.5^2+2.5}{1-5}=-15\\B=\frac{2.\left(-1\right)^2+2\left(-1\right)}{1-\left(-1\right)}=0\end{matrix}\right.\)

25 tháng 12 2020

ko có biết

27 tháng 7 2023

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

27 tháng 7 2023

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)