cho biểu thức:A = 3 mũ 1 + 3 mũ 2+3 mũ 3+.......................+3 mũ 2006
a, rút gọn A
b, tìm x để 2A + 3 =3 mũ x
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Lời giải :
1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)
\(=\frac{a^3}{4}+3ab^2\)
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...
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
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
Bài 2:
a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
Bạn viết rõ hơn nhé :
\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)
= \(\frac{x^4-xy^3}{2xy+y^2}.\frac{2x+y}{x^3+x^2y+xy^2}\)
= \(\frac{x.\left(x-y\right).\left(x^2+xy+y^2\right).\left(2x+y\right)}{y.\left(2x+y\right).x.\left(x^2+xy+y^2\right)}\)
= \(\frac{x-y}{y}\)
Chúc bạn học tốt !!!
a, A=31+32+33+...+32006
3A=32+33+...+32006+32007
3A-A=(32+33+...+32006+32007)-(31+32+33+...+32006)
2A=32007-3
A=(32007-3)/2
b, 2A=32007-3
2A+3=32007
Hay 3x=32007
=>x=2007