a) \(142\cdot4^2-\left(7\cdot5^2+4x\right)=\left(4^3\cdot157\right):4^2\)

\(\Leftrightarrow142\cdot16-\left(7\cdot25+4x\right)=\left(64\cdot157\right):16\)

\(\Leftrightarrow2272-\left(175+4x\right)=10048:16\)

\(\Leftrightarrow2272-145-4x=628\)

\(\Leftrightarrow2097-4x=628\)

\(\Leftrightarrow-4x=-1469\)

\(\Leftrightarrow x=\dfrac{1469}{4}\)

Vậy \(x=\dfrac{1469}{4}\)

b) \(\left[\left(7x+23\right):6^2-5^2\right]:29=18^2\)

\(\Leftrightarrow\left(\dfrac{7x+23}{6^2}-25\right)\cdot\dfrac{1}{29}=18\)

\(\Leftrightarrow\left(\dfrac{7x+23}{36}-25\right)\cdot\dfrac{1}{29}=18\)

\(\Leftrightarrow\dfrac{7x+23-900}{36}\cdot\dfrac{1}{29}=18\)

\(\Leftrightarrow\dfrac{7x-877}{36}\cdot\dfrac{1}{29}=18\)

\(\Leftrightarrow\dfrac{7x-877}{1044}=18\)

\(\Leftrightarrow7x-877=18792\)

\(\Leftrightarrow7x=18792+877\)

\(\Leftrightarrow7x=19669\)

\(\Leftrightarrow x=\dfrac{19669}{7}\)

a: \(\Leftrightarrow142\cdot16-7\cdot25-4x=4\cdot157\)

\(\Leftrightarrow2097-4x=628\)

=>4x=1469

hay x=1469/4

b: \(\Leftrightarrow\left[\left(7x+23\right):36-2704\right]=324\cdot29=9396\)

\(\Leftrightarrow\left(7x+23\right):36=12100\)

=>7x+23=435600

=>7x=435577

hay x=435577/7

       25-{14-[(2-x)-(x+13)+23]}=62-{8-[(2x-10)-(29-3x)+8]}

<=>25-14+2-x-x-13+23=62-8+2x-10-29+3x+8

<=>13-x-x-13+23=2x+3x+8-10-29

<=>-x-x+23=2x+3x+(-31)

<=>23-(x+x)=2x+2x-31

<=>23-2x=2x+3x-31

<=>23+31=2x+2x+3x

<=>54=x(2+2+3)

<=>54=x.7

=>x=54/7

Vậy x=54/7

<=>54=x.7

a) 142.16-(7.25+4x) = (64.157) :16

  142.16-(7.25+4x)=10048:16

142.16-(7.25+4x)=628

2272-(175+4x)=628

175+4x=2272-628

175+4x=1644

4x=1644-175

4x=1469

x=1469:4

x=367,25

b, [(7x+23):36-25]:29=324

(7x+23):36-25=324.29

(7x+23):36-25=9396

(7x+23):36=9396+25

(7x+23):36=9421

7x+23=9421.36

7x+23=339156

7x=339156-23

7x=339133

x=339133:7

x=48447,57143

a) \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

\(=>\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)

\(=>\left(4x+2\right)\left(2x-4\right)=0\)

\(=>4\left(2x+1\right)\left(x-2\right)=0\)

\(=>\orbr{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=2\end{cases}}\)

b)\(x^3-\frac{x}{49}=0=>x\left(x^2-\frac{1}{49}\right)=0=>x\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)

\(=>x=0\)hoặc \(x=\frac{1}{7}\) hoặc \(x=-\frac{1}{7}\)

a)\(\(\left(3x-1\right)^2-\left(x+3\right)^2=0\)\)

\(\(\Leftrightarrow\left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\)\)

\(\(\Leftrightarrow\left(2x-4\right)\left(4x+2\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\4x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}}\)\)

b)\(\(x^3-\frac{x}{49}=0\)\)

\(\(\Leftrightarrow\frac{49x^3-x}{49}=0\)\)

\(\(\Leftrightarrow x\left(49x^2-1\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\49x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(7x-1\right)\left(7x+1\right)=0\end{cases}}}\)\)\

\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7};x=-\frac{1}{7}\end{cases}}\)\)

c)\(\(x^2-7x+12=0\)\)

\(\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)\)

d) \(\(4x^2-3x-1=0\)\)

\(\(\Leftrightarrow4x^2-4x+x-1=0\)\)

\(\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)\)

\(\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)\)

e) Tham khảo tại : [Toán 8]Giải phương trình | Cộng đồng học sinh Việt Nam - HOCMAI Forum

https://diendan.hocmai.vn/threads/toan-8-giai-phuong-trinh.290061/

_Y nguyệt_

a thiếu đề bạn nhé

b) \(x^3-\frac{x}{49}=0\) 

\(\Rightarrow x\left(x^2-\frac{1}{49}\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{49}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\) 

Vậy.........

c)  \(x^2-7x++12=0\)  

\(\Rightarrow\left(x-3,5\right)^2-0,5^2=0\) 

\(\Rightarrow\left(x-3,5+0,5\right)\left(x-3,5-0,5\right)=0\) 

\(\Rightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\) 

Vậy.....

d) \(4x^2-3x-1=0\) 

\(\Rightarrow4x^2-3x+0,5625-1,5625=0\)

\(\Rightarrow\left(2x-0,75\right)^2-1,25^2=0\) 

\(\Rightarrow\left(2x-0,75+1,25\right)\left(2x-0,75-1,25\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}2x+0,5=0\\2x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-0,25\\x=1\end{cases}}}\) 

Vậy.....

182 x 29 + 182 x 45 - 182 x 4 + 182 x 30

= 182 x ( 29 + 45 - 4 + 30 )

= 182 x 100

= 18200

nếu mk đúng mk nhé thanks mn nhìu ạ kb đi

 = 182 x (29+45-4+30)

 = 182 x 100 = 18200

k mk nha

182 x 29 + 182 x 45 - 182 x 4 + 182 x 30

= 182 x ( 29 + 45 - 4 + 30)

= 182 x (74 - 34)

= 182 x 40

=       7280