Mỗi câu đúng được 0,75 điểm

a) x × 7 + 85 = 106

             x × 7 = 106 – 85

             x × 7 = 21

                   x = 21 : 7

                   x = 3

b) x : 9 + 72 = 137

              x : 9 = 137 - 72

              x : 9 = 65

                   x = 65 x 9

                   x = 585

con cai nit

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

a, (x + 30) – 75 = 125

=> x + 30 = 125 + 75 = 200

=> x = 200 – 30

=> x = 170

Vậy x = 170

b, x – 72 : 36 = 18

=> x – 2 = 18

=> x = 18 + 2 = 20

Vậy x = 20

c, x – 17 = 54

=> x = 54 +17

=> x = 71.

Vậy x = 71

d, 36 – (x – 2) = 12

=> x – 2 = 36 – 12

=> x = 24 + 2 = 26

Vậy x = 26

e, 9x – 7 = 837

=>9x = 837 + 7 = 844

=> x =  844 9

Vậy x =  844 9

f, (x – 15) – 107 = 0

=> x – 15 = 107

=> x = 107 +15

=> x = 122.

Vậy x = 122

g, 134 + (116 – x) = 145

=> 116 – x = 145 – 134

=> x = 116 – 11

=> x = 5.

Vậy x = 5

Bài 4:

a: =>20x=7078

=>x=353,9

b: =>x=78*11=858

Câu 5:

Chiều rộng là (96-24)/2=72/2=36

Chiều dài là 36+24=60

Diện tích là 36*60=2160m²

Số cây trồng được là:

2160:12*3=540 cây

\(a,8x-75=5x+21\)

    \(8x-5x=21+75\)

               \(3x=96\)

                  \(x=32\)

\(b,9x+25=-\left(2x-58\right)\)

    \(9x+25=-2x+58\)

    \(9x+2x=58-25\)

            \(11x=33\)

                 \(x=3\)

\(c,\left(5-x\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}}\)

a) \(8x-75=5x+21\)

\(8x-5x=21+75\)

\(3x=96\)

\(x=32\)

vậy \(x=32\)

b) \(9x+25=-\left(2x-58\right)\)

\(9x+25=-2x+58\)

\(9x+2x=58-25\)

\(11x=33\)

\(x=3\)

vậy \(x=3\)

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

\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)

vậy....

d) \(3.\left|x+3\right|-4=-7-\left(-18\right)\)

\(3.\left|x+3\right|-4=-7+18\)

\(3.\left|x+3\right|=11+4\)

\(3.\left|x+3\right|=15\)

\(\left|x+3\right|=5\)

\(\Rightarrow\orbr{\begin{cases}x+3=5\\x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-8\end{cases}}\)

vậy..

a) x – 258 = 347

               x = 347 + 258

               x = 605

b) x × 9 = 819

           x = 819 : 9

           x = 91

co ai biet ko

274 - ( 9x+18)=4

9x+18=274-4

9x+18=270

9x=270-18

9x=252

x=252:9

x=28

442+(418-x)=860

418-x=860-442

418-x=418 

x=418-418

x=0

`274 - ( 9xx x+18)=4`

`9xx x+18 =274-4`

`9xx x+18 =270`

`9 xx x=270-18`

`9 xx x=252`

`x=252:9`

`x= 28`

__

`442+(418-x)=860`

`418-x=860-442`

`418-x=418`

`x=418-418`

`x=0`

`#3107`

`274 - (9x + 18) = 4`

`=> 9x + 18 = 274 - 4`

`=> 9x + 18 = 270`

`=> 9x = 270 - 18`

`=> 9x = 252`

`=> x = 252 \div 9`

`=> x = 28`

Vậy, `x = 28.`

____

`442 + (418 - x) = 860`

`=> 418 - x = 860 - 442`

`=> 418 - x = 418`

`=> x = 418 - 418`

`=> x = 0`

Vậy, `x = 0.`

\(\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4\sqrt{x}}+\frac{4x\sqrt{x}+4\sqrt{x}}{4x^2+9x+18\sqrt{x}+9}-2=\frac{\left(-4x\sqrt{x}+4x^2+9x+22\sqrt{x}+9\right)^2}{\left(4x^2+9x+18\sqrt{x}+9\right)\left(4x\sqrt{x}+4\sqrt{x}\right)}\ge0\)

Đặt \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}\left(x>0\right)\Rightarrow M>0\)

Đặt \(y=\sqrt{x}>0\)ta có \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}=\frac{4y^4+9y^2+18y+9}{4y^3+4y^2}\)\(=\frac{3\left(4y^3+4y^2\right)+\left(4y^2-12y^3-3y^2+18y+9\right)}{4y^3+4y^2}=3+\frac{\left(2y^2-3y-3\right)^2}{4y^3+4y^2}\ge3\)

\(y>0\Rightarrow\hept{\begin{cases}4y^3+4y^2>0\\\left(2y^2-3y-3\right)^2\ge0\end{cases}\Rightarrow\frac{\left(2y-3y-3\right)^2}{4y^3+4y^2}\ge0}\)

Đẳng thức xảy ra \(\Leftrightarrow2y^2-3y-3=0\Leftrightarrow y=\frac{3+\sqrt{33}}{4}\left(y>0\right)\)

\(\Rightarrow x=\left(\frac{3+\sqrt{33}}{4}\right)^2=\frac{21+3\sqrt{33}}{8}\)

Khi đó \(A=M+\frac{1}{M}=\frac{8M}{9}+\left(\frac{M}{9}+\frac{1}{M}\right)\ge\frac{8\cdot3}{9}+2\sqrt{\frac{M}{9}\cdot\frac{1}{M}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}M=3\\\frac{M}{9}=\frac{1}{M}\end{cases}\Leftrightarrow M=3\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}}\)

Vậy \(A_{min}=\frac{10}{3}\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}\)