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a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
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: $x=\dfrac15\ \text{hoặc}\ x=\dfrac{-39+\sqrt{1481}}4\ \text{hoặc}\ x=\dfrac{-39-\sqrt{1481}}4.$
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
=> x + 1 + x + 2 + x + 3 + ... + x + 9 + x + 10 = 5
= > ( x+x+x+ .... + x ) + ( 1+2+3+4+5+6+7+8+9+ 10 )= 5
=> x . 10 + 55 = 5
=> x . 10 = -50
=> x = ( - 50 ) : 10 = -5