Cần gấp vote 5 sao
Giải pt sau
1/x(x+1) + 1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) = 1/35
(2x-3)(10x+1) < 5x(4x-2)
Cần gấp vote 5 sao
Giải pt sau
1/x(x+1) + 1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) = 1/35
(2x-3)(10x+1) < 5x(4x-2)
Mình trình bày trong hình !
Đáp án:
$a){\left[\begin{aligned}x=10\\x=-14\end{aligned}\right.}\\
b) x>\frac{-1}{6}\\$
Giải thích các bước giải:
$a) \frac{1}{x(x+1)}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}=\frac{1}{35}\\
đk:{\left\{\begin{aligned}x\neq 0\\x\neq 1\\ x\neq 2\\ x\neq 3\\ x\neq 4\end{aligned}\right.}\\
\Rightarrow \frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}=\frac{1}{35}\\
\Leftrightarrow \frac{1}{x}-\frac{1}{x+4}=\frac{1}{35}\\
\Leftrightarrow \frac{35(x+4)}{35x}-\frac{35x}{35(x+4)}=\frac{x(x+4)}{35}\\
\Leftrightarrow 35x+140-35x=x^2+4x\\
\Leftrightarrow x^2+4x-140=0\\
\Leftrightarrow x^2+14x-10x-140=0\\
\Leftrightarrow x(x+14)-10(x+14)=0\\
\Leftrightarrow (x-10)(x+14)=0\\
\Leftrightarrow {\left[\begin{aligned}x=10\\x=-14\end{aligned}\right.}\\
b) (2x-3)(10x+1)<5x(4x-2)\\
\Leftrightarrow 20x^2+2x-30x-3<20x^2-10x\\
\Leftrightarrow -28x+10x-3<0\\
\Leftrightarrow -18x<3\\
\Leftrightarrow x>\frac{-1}{6}\\$