\(=25\times\left(2+3-3\right)\)

\(=25\times2\)

\(=50\)

\(25\left(2+3-3\right)\)

\(=25\cdot2\)

\(=50\)

a: =25(15+45*3)

=25*150

=3750

b: \(=-10\left(25+75-50\right)=-10\cdot50=-500\)

c: =>3^x-2=27

=>x-2=3

=>x=5

d: =>2x-5=-4

=>2x=1

=>x=1/2

e: =>2(x-1)^2=32

=>(x-1)^2=16

=>x-1=4 hoặc x-1=-4

=>x=-3 hoặc x=5

f:  =>25(x+3)=75

=>x+3=3

=>x=0

Ai biết chỉ mình nha 

1. 11(x-9)=77

x-9=7

x=16

2. 4(x-3)=48

x-3=12

x=15

3.3(x-8)=81

(x-8)=27

x=35

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

\(=\dfrac{-1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)

\(\sqrt{25\left(x-3\right)}-10\sqrt{\dfrac{x-3}{25}}-1=3+\sqrt{x-3}\left(đk:x\ge3\right)\)

\(\Leftrightarrow5\sqrt{x-3}-10.\dfrac{1}{5}\sqrt{x-3}-1=3+\sqrt{x-3}\)

\(\Leftrightarrow2\sqrt{x-3}=4\Leftrightarrow\sqrt{x-3}=2\)

\(\Leftrightarrow x-3=4\Leftrightarrow x=7\)

Ta có: \(\sqrt{25\left(x-3\right)}-10\sqrt{\dfrac{x-3}{25}}-1=3+\sqrt{x-3}\)

\(\Leftrightarrow5\sqrt{x-3}-2\sqrt{x-3}-\sqrt{x-3}=4\)

\(\Leftrightarrow2\sqrt{x-3}=4\)

\(\Leftrightarrow x-3=4\)

hay x=7

Đặt: \(\sqrt[3]{25-x^3}=t\Leftrightarrow t^3+x^3=25\Leftrightarrow\left(t+x\right)^3-3tx\left(t+x\right)=25\)(1)

pt trở thành: 

\(xt\left(x+t\right)=30\) Thế vào (1) ta có:

\(\left(t+x\right)^3-3.30=25\)

<=> \(t+x=\sqrt[3]{115}\)

=> \(xt=\frac{30}{\sqrt[3]{115}}\)

x, t là nghiệm của phương trình bậc 2:

 \(X^2-\sqrt[3]{115}X+\frac{30}{\sqrt[3]{115}}=0\)(1)

Đen ta <0 

=> Phương trình (1) vô nghiệm.

=> Không tồn tại x

Vậy phương trình ban đầu vô nghiệm.

Đề có sai không bạn?