a,Đề bài....

=>12x-33=3^2020-2019

=>12x-33=3

=>12x=36

=>x=3

b,Đề bài.....

=>2x-3=7/3

=>2x=7/3+3

=>2x=16/3

=>x=8/3

c,Đề bài.....

=>(2x-2⁵)=8⁵:8³

=>2x-32=8^5-3

=>2x-32=64

=>2x=96

=>x=48

d,Đề bài...

=>(6x-72):2=285

=>(6x-72)=142,5

=>6x=212,5

=>x=.......

Học\(tut\)

a) (12x-33)=3                             b) 2x-3=21                    c) 2x-2^5=64                  d)   (6x-72):2=285

      12x=36                                       2x=24                             2x=96                                   6x-72=570

       x=3                                              x=12                               x=48                                     x=107

\(A=\left|-x+8\right|-21\)

\(A=\left|-x+8\right|-21\ge-21\)

\(MinA=-21\Leftrightarrow-x+8=0\)\(\Leftrightarrow x=8\)

\(B=\left|-x-17\right|+\left|y-36\right|+12\)

\(B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)

\(MinB=12\Leftrightarrow\hept{\begin{cases}-x-17=0\\y-36=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-17\\y=36\end{cases}}\)

\(C=-\left|2x+8\right|-35\)

\(C=-\left|2x+8\right|-35\le-35\)

\(MaxC=-35\Leftrightarrow2x+8=0\Leftrightarrow x=-4\)

Trl

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Học tốt 

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\(\text{a) A = | -x + 8| - 21}\)
Vì | -x + 8| \(\le\) 0 ( với mọi x )
=> A = | -x + 8| - 21\(\ge\) -21
=> Amax = -21 khi | -x + 8| = 0 => -x + 8 = 0 => -x = -8 => x = 8
Vậy với Amin = -21 thì x = 8
b) \(B=\left|-x-17\right|+\left|y-36\right|+12\)
Vì \(\left\{\begin{matrix}\left|-x-17\right|\ge0\\\left|y-36\right|\ge0\end{matrix}\right.\)=> \(\left|-x-17\right|+\left|y-36\right|\ge0\)
=> \(B=\left|-x-17\right|+\left|y-36\right|+12\le12\)
=> Bmin = 12 khi \(\left|-x-17\right|+\left|y-36\right|=0\)
=> \(\left\{\begin{matrix}\left|-x-17\right|=0\\\left|y-36\right|=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}-x-17=0\\y-36=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}-x=17\\y=36\end{matrix}\right.\)=>\(\left\{\begin{matrix}x=-17\\y=36\end{matrix}\right.\)
Vậy Bmin = 12 khi \(\left\{\begin{matrix}x=-17\\y=36\end{matrix}\right.\)
c) \(C=-\left|2x-8\right|-35\)
Vì \(-\left|2x-8\right|\ge0\)
=> \(C=-\left|2x-8\right|-35\ge-35\)
=> Cmin = -35 khi \(-\left|2x-8\right|=0\)=> \(-2x-8=0\)=>\(-2x=8\)=> \(x=4\)
Vậy Cmin = -35 khi x = 4
d) \(D=3\left(3x-12\right)^2-37\)
Vì \(\left(3x-12\right)^2\ge0\)
=> \(3\left(3x-12\right)^2\ge0\)
=> \(D=3\left(3x-12\right)^2-37\ge-37\)
=> Dmin = -37 khi \(3\left(3x-12\right)^2=0\) => \(\left(3x-12\right)^2=0\)=> \(3x-12=0\)=> \(3x=12\)=>\(x=4\)
Vậy Dmin = -37 khi x = 4

a, A=|-x+8|-21

Vì |-x+8|>hoặc =0 với mọi x

suy ra |-x+8|-21>hoặc = -21

Dấu = xảy ra khi và chỉ khi |-x+8|=0

Khi và chỉ khi -x+8=0

Khi và chỉ khi-x=-8

khi và chỉ khi x =8

Vậy GTNN của A là -21 tại x=8

\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)

\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)

\(\Leftrightarrow\left(19x+50\right):14=9\)

\(\Leftrightarrow19x+50=126\)

\(\Leftrightarrow19x=76\Leftrightarrow x=4\)

b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240

x + x + 1 + x + 2 + ... + x + 30 = 1240

( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240

Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )

Tổng là : ( 30 + 1 ) . 30 : 2 = 465

=> 31x + 465 = 1240

=> 31x = 775

=> x = 25

Vậy........

Trl:

a) \(x-9=5.8\)

\(\Rightarrow x-9=40\)

\(\Rightarrow x=49\)

Vậy \(x=49\)

b) \(x+8=-4+11\)

\(\Rightarrow x+8=7\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

c) \(\left|x\right|-2=-3.\left(-8\right)\)

\(\Rightarrow\left|x\right|-2=24\)

\(\Rightarrow\left|x\right|=26\)

\(\Rightarrow x=-26\)

\(\Rightarrow\orbr{\begin{cases}x=26\\x=-26\end{cases}}\)

Vậy \(x\in\left\{26;-26\right\}\)

d) \(\left|x-2\right|=\left|-72\right|\)

\(\Rightarrow\left|x-2\right|=72\)

\(\Rightarrow\orbr{\begin{cases}x-2=72\\x-2=-72\end{cases}\Rightarrow}\orbr{\begin{cases}x=72+2\\x=-72+2\end{cases}\Rightarrow\orbr{\begin{cases}x=74\\x=70\end{cases}}}\)

Vậy \(x\in\left\{74;70\right\}\)

Câu a:

- \(\frac12\)(3 - 2\(x\)) - 7 = 5 - \(\frac13\)(\(x\) - \(\frac45\))

- \(\frac32\) + \(x\) - 7 = 5 - \(\frac13x\) + \(\frac{4}{15}\)

\(x\) + \(\frac13x\) = 5 + \(\frac{4}{15}\) + 7 + \(\frac32\)

(1 + \(\frac13\))\(x\) = \(\frac{150}{30}\) + \(\frac{8}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{158}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{368}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{413}{30}\)

\(x\) = \(\frac{413}{30}\) : \(\frac43\)

\(x\) = \(\frac{413}{40}\)

Vậy \(x=\frac{413}{40}\)

Câu b:

(5 - 3x/2) : - 1 3/8 = - 7 1/3

(5 - 3x/2) : (-11/8) = - 22/3

5 - 3x/2 = - 22/3 x (-11/8)

5 - 3x/2 = 121/12

3x/2 = 5 - 121/12

3x/2 = - 61/12

x = - 61/12 : 3/2

x = -61/18

Vậy x = - 61/18

a, \(\left(2600+6400\right)-3x=1200\)

\(\Rightarrow9000-3x=1200\)

\(\Rightarrow3x=7800\)

\(\Rightarrow x=2600\)

b, \(\left[\left(6x-72\right):2-84\right].28=5628\)

\(\Rightarrow\left(6x-72\right):2-84=201\)

\(\Rightarrow\left(6x-72\right):2=285\)

\(\Rightarrow6x-72=570\)

\(\Rightarrow6x=642\)

\(\Rightarrow x=107\)

a) (2600 + 6400) - 3x = 1200

9000 - 3x = 1200

3x = 9000 - 1200

3x = 7800

x = 2600

Vậy x = 2600

b) [(6x - 72) : 2 - 48].28 = 5628

(6x - 72) : 2 - 48 = 5628 : 28

(6x - 72) : 2 - 48 = 201

(6x - 72) : 2 = 201 + 48

(6x - 72) : 2 = 249

6x - 72 = 249.2

6x - 72 = 498

6x = 498 + 72

6x = 570

x = 570 : 6

x = 95

Vậy x = 95