a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

phep 1=49

phep 2 hinh nhu sai de

phep 3 lay(34571-34567)/2

phep 4 =505

tinh bieu thuc=67

(23x7)+78=X+190

161+78=X+190

239=x+190

x=239-190

x=49

(23*7) + 78 = x + 190 

161 + 78 = x + 190 

239 = x + 190 

239 - 190 = x 

x = 49

Đề:......

<=> 7. (x + 25) = 210

<=> x + 25 = 210 : 7

<=> x + 25 = 30

<=> x = 5

Vậy x = 5

\(7\left(x+5^2\right)-20=190\)

\(7\left(x+25\right)=190+20\)

\(7\left(x+25\right)=210\)

\(x+25=210:7\)

\(x+25=30\)

\(x=30-25\)

\(x=5\)

Vậy \(x=5\)

A, 7[x + 5] - 20 = 190

7x + 35 - 20 = 190

7x + 15 = 190

7x = 175

x = 25

B, 155 - 10[x + 1] = 55

155 - 10x - 10 = 55

-10x + 90 = 55

-10x = -35

x = 3.5

C, 6[x + 2^3] + 40 = 100

6[x + 8] + 40 = 100

6x + 48 + 40 = 100

6x + 88 = 100

6x = 1

2 x = 2

D, 15x - 133 = 17

15x = 150

x = 10

E, 90[x + 2] = 45

90x + 180 = 45

90x = -135

x = -1.5

F, 4x + 54 = 82

4x = 28

x = 7

G, 17x - 20 = 14

17x = 34

x = 2

7 \(\times\) ( 2\(x\) - 5) - 5 \(\times\) (7\(x\) - 2) + 2 \(\times\) (5\(x\) - 7) = (\(x\) - 2) - (\(x\) +4)

14\(x\) - 35 - 35\(x\) + 10 + 10\(x\) - 14 = \(x\) - 2 - \(x\) - 4

(14\(x\) - 35\(x\) + 10\(x\)) - (35 - 10+ 14)  = -6

(- 21 \(x\) + 10\(x\)) - (25 + 14) = - 6

         -11\(x\) - 39 = - 6

          -11\(x\)       = - 6 + 39

          - 11\(x\)     = 33

                \(x\) = 33 : (-11)

                \(x\) =  - 3

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