128-3*(x+4)=23

3*(x+4)=128-23=105

x+4=105/3=35

x=31

`(13+x)/20 = 3/4`

`(13+x) xx4=3xx20`

`(13+x)xx4=60`

`13+x=60:4`

`13+x=15`

`x=15-13`

`x=2`

__

`(23-x)/25 =4/5`

`(23-x)xx5=4xx25`

`(23-x)xx5=100`

`23-x=100:5`

`23-x=20`

`x=23-20`

`x=3`

a) x-7 = - 5

<=> x= 2

b) 128 - 3.x.(x+4) = 23

<=> - 3x² - 12x +105=0

<=> -x² -4x + 35 =0

\(\Delta'=156\Rightarrow\sqrt{\Delta'}=2\sqrt{39}\)

=> \(\left[{}\begin{matrix}x=-2+\sqrt{39}\\x=-2-\sqrt{39}\end{matrix}\right.\)

vậy S= \(\left\{-2+\sqrt{39};-2-\sqrt{39}\right\}\)

( 51 x 52 - 49 x 48 ) x ( 45 x 128 ) x ( 45 x128 - 90 x 64) x ( 1995 x 1996 x 1997)

= ( 51 x 52 - 49 x 48 ) x ( 45 x 128 ) x ( 5760 - 5760 ) x ( 1995 x 1996 x 1997)

= ( 51 x 52 - 49 x 48 ) x ( 45 x 128 ) x 0 x ( 1995 x 1996 x 1997)

= 0.

#Y/n

Tìm x thuoc z:

1) \(26-\left|x+9\right|=-13\)

\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)

\(\Leftrightarrow\left|x+9\right|=39\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)

Vậy: \(x\in\left\{30;-48\right\}\)

2) \(\left|x+7\right|-13=25\)

\(\Leftrightarrow\left|x+7\right|=25+13=38\)

\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)

\(\Leftrightarrow x\in\left\{31;-45\right\}\)

Vậy:.................

tim x biet

\(1)123-3.\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=123-23\)

\(\Leftrightarrow3\left(x+4\right)=100\)

\(\Leftrightarrow x+4=\frac{100}{3}\)

\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)

Vậy:................

2) Tương tự

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