\(x:\dfrac{1}{7}=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{1}{4}.\dfrac{1}{7}\\ \Rightarrow x=\dfrac{1.1}{4.7}\\ \Rightarrow x=\dfrac{1}{28}\)

x:17=14⇒x=14.17⇒x=1.14.7x=128

\(x:\dfrac{1}{7}=14\)

<=> \(\dfrac{1}{7x}=14\)

ĐKXĐ: x\(\ne\)0

<=> 1 = 14 . 7x

<=> 98x =1

<=> x = \(\dfrac{1}{98}\)

\(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Rightarrow\left(x^3+2x^2-5x-10\right)+5x=2x^2+17\)

\(\Rightarrow x^3+2x^2-5x-10+5x=2x^2+17\)

\(\Rightarrow x^3+2x^2-10=2x^2+17\)

\(\Rightarrow x^3-10=17\)

\(\Rightarrow x^3=17+10=27\)

\(\Rightarrow x^3=3^3\)

\(\Rightarrow x=3\)

(x2−5)(x+2)+5x=2x2+17

⇒(x3+2x2−5x−10)+5x=2x2+17

⇒x3+2x2−5x−10+5x=2x2+17

⇒x3+2x2−10=2x2+17

⇒x3−10=17

⇒x3=17+10=27

⇒x3=33

⇒x=3

\(-2x^2+5x=16\)

\(-2x^2+5x-16=0\)

\(-\left(2x^2-5x+16\right)=0\)

\(2x^2-5x+16=0\)

\(2\left(x^2-\frac{5}{2}x+8\right)=0\)

\(x^2-\frac{5}{2}x+8=0\)

\(x^2-\frac{5}{2}x+\frac{25}{16}+\frac{103}{16}=0\)

\(\left(x-\frac{5}{4}\right)^2+\frac{103}{16}=0\)

Ta có: \(\left(x-\frac{5}{4}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{5}{4}\right)^2+\frac{103}{16}\ge\frac{103}{16}>0\)

Mà: \(\left(x-\frac{5}{4}\right)^2+\frac{103}{16}=0\)

=> Vô lí

Vậy : ko có giá trị thỏa mãn của x 

=.= hok tốt!!

( Tham khảo nhé  !)

Trả lời :

( Ảnh dưới )

ai biet

\(x.[\left(-2x\right)^2+5x]=16\)

\(\Rightarrow x.[\left(-4x\right)+5x]=16\)

\(\Rightarrow x.x=16\)

\(\Rightarrow2x=16\Rightarrow x=8\)

Vay \(x=8\)

A = |2x - 5| + 3 - 2x

A = 2x - 5 + 3 - 2x

A = (2x - 2x) + (-5 + 3)

A = -2

B = |x² - 5x + 4| - 4 + 5x - x²

B = x² - 5x + 4 - 4 + 5x - x²

B = (x² - x²) + (-5x + 5x) + (4 - 4)

B = 0