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Ta có: \(\left(a^{100}+b^{100}\right)\cdot ab=a^{101}\cdot b+b^{101}\cdot a\)
\(\left(a^{101}+b^{101}\right)\cdot\left(a+b\right)=a^{102}+a^{101}\cdot b+b^{101}\cdot a+b^{102}\)
Do đó: \(\left(a^{101}+b^{101}\right)\left(a+b\right)-\left(a^{100}+b^{100}\right)\cdot ab\)
\(=a^{102}+b\cdot a^{101}+a\cdot b^{101}+b^{102}-a^{101}\cdot b-b^{101}\cdot a\)
\(=a^{102}+b^{102}\)
Kết hợp đề bài, ta có:
\(\left(a^{102}+b^{102}\right)\left(a+b\right)-\left(a^{102}+b^{102}\right)\cdot ab=a^{102}+b^{102}\)
\(\Leftrightarrow a+b-ab=1\)
\(\Leftrightarrow a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)+b\left(1-a\right)=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(1-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\1-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
Vậy: \(P=a^{2004}+b^{2004}=1^{2004}+1^{2004}=2\)
Ta có :
\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
Do \(a+b=a^3+b^3\)
\(\Rightarrow a+b=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\Rightarrow a^2-ab+b^2=1\)
Mà \(a^2=b^2=a+b\) ,ta có :
\(a+b-ab=1\)
\(\Rightarrow a+b-ab-1=0\)
\(\Rightarrow\left(a-1\right)-\left(ab-b\right)=0\)
\(\Rightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Rightarrow\left(a-1\right)\left(1-b\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}a-1=0\\1-b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
Thay vaò biểu thức ,có :
\(1^{2015}+1^{2015}=1+1=2\)
a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)
= x³ - 125 - x² + 4 + x³ + x² + 4x
= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)
= 2x³ + 4x - 121
b) Tại x = -2 ta có:
A = 2.(-2)³ + 4.(-2) - 121
= 2.(-8) - 8 - 121
= -16 - 129
= -145
c) x² - 1 = 0
x² = 1
x = -1; x = 1
*) Tại x = -1 ta có:
A = 2.(-1)³ + 4.(-1) - 121
= 2.(-1) - 4 - 121
= -2 - 125
= -127
*) Tại x = 1 ta có:
A = 2.1³ + 4.1 - 121
= 2.1 + 4 - 121
= 2 - 117
= -115
a) M có nghĩa khi a 3 - 4 a ≠ 0 ⇔ a ≠ { 0 ; ± 2 }
b) Rút gọn thu được: M = a ( a 2 + 4 a + 4 ) a ( a 2 − 4 ) = a + 2 a − 2
c) M = − 3 ⇔ a + 2 a − 2 = − 3 ⇔ a = 1 (TMĐK)
a) a ≠ 0 , a ≠ − 5
b) Ta có A = a 3 + 4 a 2 − 5 a 2 a ( a + 5 ) = a ( a − 1 ) ( a + 5 ) 2 a ( a + 5 ) = a − 1 2
c) Thay a = -1 (TMĐK) vào a ta được A = -1
d) Ta có A = 0 Û a = 1 (TMĐK)
Câu 2:
a: ĐKXĐ: \(x\notin\left\{0;2\right\}\)
b: Sửa đề: \(A=\left(\dfrac{2x-x^2}{2x^2+8}-\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\left(\dfrac{2}{x^2}-\dfrac{x-1}{x}\right)\)
\(=\left(\dfrac{2x-x^2}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{2-x\left(x-1\right)}{x^2}\)
\(=\left(\dfrac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{2-x^2+x}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)\left(x+1\right)}{2\left(x^2+4\right)\cdot x^2}=\dfrac{x+1}{2x}\)
c: Khi x=2024 thì \(A=\dfrac{2024+1}{2\cdot2024}=\dfrac{2025}{4048}\)
Câu 1:
a: \(25x^2\left(x-3y\right)-15\left(3y-x\right)\)
\(=25x^2\left(x-3y\right)+15\left(x-3y\right)\)
\(=\left(x-3y\right)\left(25x^2+15\right)\)
\(=\left(x-3y\right)\cdot5\cdot\left(5x^2+3\right)\)
b: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)