Bài 1 : Rút gọn biểu thức sau :
a) A = x + \(|x|\)
b) B = 2. ( 3x - 1) - \(|5-x|\)
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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
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)
b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)
Khi x=-1/2 thì B=2/5
c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
a, đk : x khác -2 ; 2
\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)
b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)
Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)
Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)
c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)
2-x | 1 | -1 |
x | 1 | 3 |
a)
+) Nếu x > 0 thì A = x + x = 2x
+) Nếu x = 0 thì A = 0 + 0 = 0
+) Nếu x < 0 thì A = -x + x = 0
b) B = 2 ( 3x - 1 ) - | 5 - x |
B = 6x - 2 - | 5 - x |
Xét 3 t/h như câu a)
a)+) Nếu x > 0 thì A = x + x = 2x
+) Nếu x = 0 thì A = 0 + 0 = 0
+) Nếu x < 0 thì A = -x + x = 0
b) B = 2 ( 3x - 1 ) - | 5 - x |
B = 6x - 2 - | 5 - x |
Xét 3 t/h như câu a)