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\(2x+4x+6x+........+98x+100x=51.10^2\)

\(\Rightarrow x\left(2+4+6+.........+98+100\right)=51.100\)

\(\Rightarrow x.2550=5100\)

\(\Rightarrow x=5100:2550\)

\(\Rightarrow x=2\)

Vậy x = 2

\(2x+4x+6x+...+100x=51.10^2\)

\(\Rightarrow x\left(2+4+...+100\right)=5100\)

\(\Rightarrow x\left(\frac{\left[\left(100-2\right):2+1\right].\left(100+2\right)}{2}\right)=5100\)

\(\Rightarrow x\left(\frac{50.102}{2}\right)=5100\)

\(\Rightarrow\frac{5100x}{2}=5100\)

\(\Rightarrow5100x=5100.2\)

\(\Rightarrow x=2\)

2x + 4x + 6x + ...+ 98x + 100x = 51.10²

x.(2+4+6+...+98+100) = 51.100

\(x.\left[\left(2+100\right).50:2\right]=5100\)

x. 2550 = 5100

x = 5100 : 2550

x = 2

Phải có tổng chứ ???

P/S : ghi đề mak cg màu mè

hok tốt

a) = 5100

b) = 7500

em quên mấy pro ạ=))

(x+1)+(x+3)+.........+(x+49)=700

x+1+x+3+........+x+49=700

25x+(1+3+5+......+49)=700

25x+625=700

25x=700-625

25x=75

x=75:25

x=3

2x+4x+6x+...........+100x=5100

(2+4+6+....+100)x=5100

2550x=5100

x=5100:2550

x=2

Phải tính cả cụm luôn hả bạn

​

a)\(3x^2-8x+4\)

\(=3x^2-2x-6x+4\)

\(=x\left(3x-2\right)-2\left(3x-2\right)\)

\(=\left(x-2\right)\left(3x-2\right)\)

b)\(4x^4+81\)

\(=4x^4+36x^2+81-36x^2\)

\(=\left(2x^2+9\right)^2-36x^2\)

\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)

c)\(x^8+98x^4+1\)

\(=\left(x^8+2x^4+1\right)+96x^4\)

\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)

\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)

\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)

\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)

\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)

d)\(x^4+6x^3+7x^2-6x+1\)

\(=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)

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

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

\(A=2(\sin ^6x+\cos ^6x)-3(\sin ^4x+\cos ^4x)\)

\(=2(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)

\(=2(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)

\(=-(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x)=-(\sin ^2x+\cos ^2x)^2=-1^2=-1\)

là giá trị không phụ thuộc vào biến (đpcm)

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\(B=\sin ^6x+\cos ^6x-2\sin ^4x-\cos ^4x+\sin ^2x\)

\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-2\sin ^4x-\cos ^4x+\sin ^2x\)

\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x-2\sin ^4x-\cos ^4x+\sin ^2x\)

\(=-\sin ^4x-\sin ^2x\cos ^2x+\sin ^2x=-\sin ^2x(\sin ^2x+\cos ^2x)+\sin ^2x\)

\(=-\sin ^2x+\sin ^2x=0\)

là giá trị không phụ thuộc vào biến (đpcm)

\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)

\(=(\sin ^4x+\cos ^4x-1).\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=(\sin ^4x+\cos ^4x-1).\frac{(\sin ^2x+\cos ^2x)^2}{\sin ^2x\cos ^2x}\)

\(=(\sin ^4x+\cos ^4x-1).\frac{1}{\sin ^2x\cos ^2x}=\frac{(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}\)

\(=\frac{(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{1-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{-2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=-2\)

là giá trị không phụ thuộc vào biến $x$

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\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)

\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\cos ^6x+3\sin ^2x\cos ^2x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{\cos ^4x+2\sin ^2x}{\cos ^4x}\)

\(=1+\frac{2\sin ^2x}{\cos ^4x}\)

Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.

Ta có: \(A=\sin^6x+3\cdot\sin^4x\cdot\cos^2x+3\cdot\sin^2x\cdot\cos^4x+\cos^6x\)

\(=\left(\sin^2x+\cos^2x\right)^3\)

=1