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Bài 1 :
a ) Ta có :
\(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)
\(=15^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
b ) Ta có :
\(x=11\Rightarrow x+1=12\)
Thay \(x+1=12\) vào biểu thức ta được :
\(x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+111\)
\(=x^4-x^4-x^3+x^3-x^2+x^2-x+111\)
\(=111-x\)
Thay \(x=11\) vào biểu thức vừa rút gọn ta được :
\(111-11=100\)
\(a,3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)
\(=\left(3.5\right)^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
Bài 2:
\(a,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2.\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(b,3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
Bài 1.
a. \(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)=15^4-\left(15^4-1\right)=1\)
b. \(x=11\Rightarrow x+1=12\)
Từ đây, ta có: \(x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+111=-x+111=-11+111=100\)
Bài 2.
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
