3600 : [ ( 5x + 335 ) : x ] = 50

<=> (5x + 335 ) : x = 3600 : 50 =72 

<=> 5x + 335 = 72x 

<=> 335 = 72x - 5x 

<=> 335 = 67x 

<=> x = 335 : 67 = 5 

3600 : [ ( 5x + 335 ) : x ] = 50

<=> ( 5x + 335 ) : x = 3600 : 50 = 72

<=> 5x + 335 = 72x

<=> 335 = 72x - 5x 

<=> 335 = 67x

<=> x = 335 : 67 = 5

Cách 2: Do \(\left|x\right|\ge0\forall x\) nên \(\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)

\(\Rightarrow5x-10\ge0\Rightarrow x\ge2\)

Với \(x\ge2\), ta có : \(x+7>0;x+1>0;x-2\ge0\)

Suy ra \(x+1+x-2+x+7=5x-10\)

\(\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(tm\right)\)

Vậy x = 8.

Cách 1: Với \(x\le-7\), ta có : \(x+7\le0;x+1< 0;x-2< 0\)

Suy ra \(-x-1-x+2-x-7=5x-10\)

\(\Leftrightarrow-8x=-4\Leftrightarrow x=\frac{1}{2}\left(l\right)\)

Với \(-7< x\le-1\), ta có : \(x+7>0;x+1\le0;x-2< 0\)

Suy ra \(-x-1-x+2+x+7=5x-10\)

\(\Leftrightarrow-6x=-18\Leftrightarrow x=3\left(l\right)\)

Với \(-1< x\le2\), ta có : \(x+7>0;x+1>0;x-2\le0\)

Suy ra \(x+1-x+2+x+7=5x-10\)

\(\Leftrightarrow-6x=-20\Leftrightarrow x=\frac{10}{3}\left(l\right)\)

Với \(x>2\), ta có : \(x+7>0;x+1>0;x-2>0\)

Suy ra \(x+1+x-2+x+7=5x-10\)

\(\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(tm\right)\)

Vậy x = 8.

1) 

(=)x² = 8² + 6² = 64+36=100=10² = (-10)² 

=> x=10 hoặc x=-10

2)

(=)|x-1| = -26/-24=13/12

=> x-1 = 13/12 hoặc x-1=-13/12

=> x= 25/12 hoặc x= -1/12

3) 

(2x-4+7)\(⋮\left(x-2\right)\) 

(=) 2(x-2) + 7 \(⋮\left(x-2\right)\)

(=) 7 \(⋮\left(x-2\right)\)

(=) x-2 \(\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

(=) x\(\in\left\{-5;1;3;9\right\}\)

vì x bé nhất => x=-5

#Học-tốt

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)

\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)

\(\Rightarrow5x+6=2011\)

\(\Rightarrow5x=2011-6\)

\(\Rightarrow5x=2005\)

\(\Rightarrow x=401\)

a) Ta có \(x+4=(x+1)+3\)

nên \((x+4)\) \(⋮\left(x+1\right)\) khi \(3⋮\left(x+1\right)\) , tức là \(x+1\) là ước của 3

Vì Ư(3) = { \(-1;1;-3;3\) }

Ta có bảng

b) Ta có : \(4x+3=4(x-2)+11\)

nên \(\left(4x+3\right)⋮\left(x-2\right)\) khi \(11⋮\left(x-2\right)\) , tức là \((x-2) \) là ước của 11

( Làm tương tự thôi phần a) )

\(\Rightarrow x\in\left\{-9;1;3;13\right\}\)

b) Tìm $x$

$10(8x+9x)+8(2x-1)-2(5-6x)=20$

$\Leftrightarrow 170x+16x-8-10+12x=20$

$\Leftrightarrow 198x-18=20$

$\Leftrightarrow 198x=38$

$\Leftrightarrow x=\dfrac{19}{99}$.

Vậy: $x=\dfrac{19}{99}$.

c) Tìm $x$

$(5x^2)(194+20x)=50$

$\Leftrightarrow x^2(194+20x)=10$

$\Leftrightarrow 10x^3+97x^2-5=0$

Thử $x=\dfrac15$:

$10\left(\dfrac15\right)^3+97\left(\dfrac15\right)^2-5=0$.

Suy ra: $(5x-1)(2x^2+39x+5)=0$.

Do đó: $x=\dfrac15$ hoặc $x=\dfrac{-39\pm\sqrt{39^2-4\cdot2\cdot5}}{4}$

$=\dfrac{-39\pm\sqrt{1481}}{4}$.

Vậy: $x=\dfrac15\ \text{hoặc}\ x=\dfrac{-39+\sqrt{1481}}4\ \text{hoặc}\ x=\dfrac{-39-\sqrt{1481}}4.$

-210 = -(1 + 2 + 3 + ...+ x + 1 + x)

=> 210 = 1 + 2 + 3 + ... + x + x + 1

=> 210 = \(\frac{\left(x+1+1\right).\left(x+1\right)}{2}\)

=> 420 = (x + 2).(x + 1)

=> (19 + 2).(19 + 1) = (x + 2).(x + 1)

=> x = 19