\(\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)=192\)

\(\Leftrightarrow60k^3+86k^2+40k+6=192\)

\(\Leftrightarrow60k^3+86k^2+40k-186=0\)

\(\Leftrightarrow2\left(30k^2+73k+93\right)\left(k-1\right)=0\)

\(\Leftrightarrow k=1\)

Đề ghi thiếu rồi bạn!

đề thiếu r bn

Đặt B = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Đặt C = \(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{75.76}\)(sửa lại đề)

=> 151C = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{75.76}\)

=> 151C =\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

=> C = \(\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}{151}\)

Khi A = B : C 

= \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{151}\right)=151\)

Vậy A = 151

a, \(\left|x-3,5\right|+\left|x-\frac{1}{3}\right|=0\)

\(\hept{\begin{cases}x-3,5\ge0\forall x\\x-\frac{1}{3}\ge0\forall x\end{cases}\Rightarrow\left|x-3,5\right|+\left|x-\frac{1}{3}\right|\ge0\forall x}\)

Dấu ''='' xảy ra <=> \(x-3,5=0\Leftrightarrow x=3,5\)

\(x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\)

b, \(\left|x\right|+x=\frac{1}{3}\Leftrightarrow\left|x\right|=\frac{1}{3}-x\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-x\\x=-\frac{1}{3}+x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0\ne-\frac{1}{3}\end{cases}\Leftrightarrow}x=\frac{1}{6}}\)

c, \(\left|x-2\right|=x\Leftrightarrow\orbr{\begin{cases}x-2=x\\x-2=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}-2\ne0\\x=1\end{cases}}}\)

d, tương tự c 

Sửa ý a) của bạn @akirafake 

a) \(\left|x-3,5\right|+\left|x-1,3\right|=0\)

Ta có : \(\left|x-3,5\right|+\left|x-1,3\right|=\left|-\left(x-3,5\right)\right|+\left|x-1,3\right|=\left|3,5-x\right|+\left|x-1,3\right|\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có :

\(\left|3,5-x\right|+\left|x-1,5\right|\ge\left|3,5-x+x-1,5\right|=\left|2\right|=2\)

mà \(\left|x-3,5\right|+\left|x-1,3\right|=0\)( vô lí )

Vậy không có giá trị của x thỏa mãn 

b) \(\left|x\right|+x=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{3}-x\)

=> \(\orbr{\begin{cases}x=\frac{1}{3}-x\\x=x-\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{1}{3}\\0x=-\frac{1}{3}\end{cases}\Rightarrow}2x=\frac{1}{3}\Rightarrow x=\frac{1}{6}\)

c) \(\left|x\right|-x=\frac{3}{4}\)

=> \(\left|x\right|=\frac{3}{4}+x\)

=> \(\orbr{\begin{cases}x=\frac{3}{4}+x\\x=-x-\frac{3}{4}\end{cases}\Rightarrow}\orbr{\begin{cases}0x=\frac{3}{4}\\2x=-\frac{3}{4}\end{cases}}\Rightarrow2x=-\frac{3}{4}\Rightarrow x=-\frac{3}{8}\)

d) \(\left|x-2\right|=x\)

=> \(\orbr{\begin{cases}x-2=x\\x-2=-x\end{cases}}\Rightarrow\orbr{\begin{cases}0x=2\\2x=2\end{cases}}\Rightarrow2x=2\Rightarrow x=1\)

e) \(\left|x+2\right|=x\)

=> \(\orbr{\begin{cases}x+2=x\\x+2=-x\end{cases}}\Rightarrow\orbr{\begin{cases}0x=-2\\2x=-2\end{cases}}\Rightarrow2x=-2\Rightarrow x=-1\)

Thế x = -1 ta được :

\(\left|-1+2\right|=-1\)( vô lí )

=> Không có giá trị của x thỏa mãn

\(ab+bc-c^2-ac+1=0\)

\(< =>b\left(a+c\right)-c\left(a+c\right)+1=0\)

\(< =>\left(b-c\right)\left(a+c\right)=-1\)

\(< =>a+b=0\)

\(< =>A=\left(a+b\right)^3=0^3=0\)

không hiểu thì hỏi mình chỉ cho

Ta có ab - c² + bc - ac + 1 = 0

=> (ab + bc) - (ac + c²) + 1 = 0

=> b(a + c) -c(a + c) + 1 = 0

=> (b - c)(a + c) = - 1 (1)

Vì a;b;c nguyên

=> \(\hept{\begin{cases}b-c\inℤ\\a+c\inℤ\end{cases}}\)

Ta có -1 = (-1).1 = 1.(-1)

Khi đó (b - c)(a + c) = 1.(-1) = (-1).1

Nếu  \(\hept{\begin{cases}b-c=1\\a+c=-1\end{cases}}\Rightarrow b-c+a+c=0\Rightarrow a+b=0\)

Nếu \(\hept{\begin{cases}b-c=-1\\a+c=1\end{cases}}\Rightarrow a+c+b-c=0\Rightarrow a+b=0\)

Vậy a + b = 0

Khi đó A = 0³ = 0