(x+1)+(2x+3)+(3x+5)+...+(50x+99)=3775

=> (x + 2x + 3x + ... + 50x) + (1 + 3 + 5 + ... + 99) = 3775

=> x . (1 + 2 +3 + ...+ 50) + (1 + 3 + 5 + ... + 99) = 3775

Áp dụng công thức tính dãy số ta có :

\(1+2+3+...+50=\frac{\left[\left(50-1\right):1+1\right].\left(50+1\right)}{2}=\frac{50.51}{2}=25.51=1275\)

\(1+3+5+...+99=\frac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}=\frac{50.100}{2}=50.50=2500\)

=> x . 1275 + 2500 = 3775

=> 1275x = 1275

=> x = 1

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

\(129-50x^2=-159\)

\(\Leftrightarrow50x^2=288\)

\(\Leftrightarrow x^2=\frac{144}{25}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{12}{5}\\x=-\frac{12}{5}\end{array}\right.\)

\(129-50x^2=-159\)

\(\Rightarrow50x^2=288\)

\(\Rightarrow x^2=\frac{144}{25}\)

\(\Rightarrow x=\frac{12}{5}\) hoặc \(x=\frac{-12}{5}\)

Vậy \(x=\frac{12}{5}\) hoặc \(x=\frac{-12}{5}\)

129 - 50x² = -159

-50x² = -159 - 129

-50x² = -288

x² = \(\frac{-288}{-50}\)

x² = \(\frac{144}{25}\)

x = \(\sqrt{\frac{144}{25}}\)

x = \(\frac{12}{5}\)

( 100x - 50x - 50x ) + x + 100 = 100

=>             x + 100 = 100

=>            x = 0

Ta có: 100x + x - 50x + 100 - 50x = 100

=> (100x + x - 50 - 50x) + 100 = 100

=> x + 100 = 100

=> x = 100 - 100

=> x = 0

100x + x - 50x + 100 - 50x=100

<=>(100x-50x-50x)+x=100-100

<=>x=0

Vay x=0

bằng 1

f(x)= x^6 - 50x^5+50x^4-50x^3+50x^2-50x+50 tại x=49

<=> \(f_{\left(49\right)}\)= 49^6 - 50.49^5+50.49^4-50.49^3+50.49^2-50.49+50 

<=> \(f_{\left(49\right)}\)= 13519544083, 0396489851

kết quả =

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