d, \(\left(3x-2^4\right).7^3=2.7^4\)

\(\Rightarrow3x-2^4=2.7^4:7^3\)

\(\Rightarrow3x-16=2.7\\ \Rightarrow3x=14+16\\ \Rightarrow3x=30\Rightarrow x=10\)

Vậy.....

e, \(x-\left[42+\left(-28\right)\right]=-8\)

\(\Rightarrow x-14=-8\\ \Rightarrow x=6\)

Vậy.....

g, \(x-7=-5\)

\(\Rightarrow x=-5+7\Rightarrow x=2\)

Vậy.....

h, \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15-15+20\)

\(\Rightarrow-5x=-10\Rightarrow x=2\)

Vậy.....

Chúc bạn học tốt!!!

d/ \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

\(\Rightarrow3x-16=\dfrac{2\cdot7^4}{7^3}=14\)

\(\Rightarrow3x=14+16=30\)

\(\Rightarrow x=\dfrac{30}{3}=10\)

e/ Đễ ==> tự lm thì tốt hơn nhé

g/ Đễ ==> tự lm thì tốt hơn nhé

h/ \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15+20-15=-10\)

\(\Rightarrow x=\dfrac{-10}{-5}=2\)

i/ \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+7+25\)

\(\Rightarrow-x=0\Rightarrow x=0\)

k/ \(\left|x+2\right|=0\Rightarrow x+2=0\Rightarrow x=-2\)

l/ \(\left|x-3\right|=7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

m/ \(\left|x-5\right|=\left|-7\right|\Rightarrow\left|x-5\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)

i, \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+25+7\)

\(\Rightarrow x=0\)

Vậy.....

k, \(\left|x+2\right|=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

Vậy.....

l, \(\left|x-3\right|=-7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=-5\)

Với mọi giá trị của \(x\in Z\) ta có:

\(\left|x-3\right|\ge0\) mà \(-5< 0\) nên không tìm được giá trị nào của x thoả mãn \(\left|x-3\right|=-5\).

Vậy \(x\in\varnothing\)

m, \(\left|x-5\right|=\left|-7\right|\)

\(\Rightarrow\left|x-5\right|=7\\ \Rightarrow\left\{{}\begin{matrix}x-5=-7\\x-5=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\x=12\end{matrix}\right.\)

Vậy...,..

Chúc bạn học tốt!!!

Phần a câu viết rỗ đề cho tớ nha !

b) \(\dfrac{8}{9}\)x( 2x² - 3) = 0

*) \(\dfrac{8}{9}\)x = 0 --> x = 0

*) 2x² - 3 = 0 --> 2x² = 3 --> x² = \(\dfrac{3}{2}\)--> x = \(\sqrt{\dfrac{3}{2}}\)

Vậy .....

Bài 2:

a: =>x+32=0

=>x=-32

b: =>x-1=0

=>x=1

c: =>45-x=0 hoặc x=0

=>x=0 hoặc x=45

d: =>x-12=0 hoặc x+27=0

=>x=12 hoặc x=-27

Bài 1:

a) \(\left(m+2\right).3-5=4\)

\(\Leftrightarrow3m+6-5=4\)

\(\Leftrightarrow3m+1=4\)

\(\Leftrightarrow3m=4-1\)

\(\Leftrightarrow3m=3\)

\(\Leftrightarrow m=1\)

Vậy: m = 1

b) \(\left(m-3\right).\left(-2\right)+8=-10\)

\(\Leftrightarrow-2m+6+8=-10\)

\(\Leftrightarrow-2m+14=-10\)

\(\Leftrightarrow-2m=-10-14\)

\(\Leftrightarrow-2m=-24\)

\(\Leftrightarrow m=12\)

Vậy: m = 12

Bài 2:

a) \(\left(x-2\right)^2=9\)

\(\Leftrightarrow\left(x-2\right)^2=3^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

b) \(\left(x+3\right)^2-0,16=0\)

\(\Leftrightarrow\left(x+3\right)^2=0,16\)

\(\Leftrightarrow\left(x+3\right)^2=\left(0,4\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0,4\\x+3=-0,4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2,6\\x=-3,4\end{matrix}\right.\)

c) \(x^3=25x\)

\(\Leftrightarrow x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)

Bài 4:

a, (x-3).(7-x)=0

=> x-3=0 hoặc 7-x=0

=> x-3=0 => x=3 hoặc

7-x=0 => x=7

Vậy x=3 hoặc x=7

4b, Ta có (x+1).(x²-4)=0

=> x+1 =0 hoặc x-4=0

Th1: x+1=0

=> x=-1

Th2: x²-4=0

=> x²=4=2²

=> x=2

Vậy x=1 hoặc x=2 tại (x+1).(x²-4)=0

3.

c) \(25-\left(30+x\right)=1-\left(27-8\right)\)

\(25-30-x=1-19\)

\(-5-x=-18\)

\(x=13\)

4.

a) \(\left(x-3\right)\left(7-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

Vậy x = 3 hoặc x = 7

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

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=4=\left(\pm2\right)^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)

Vậy..

c) \(\left(x-3\right)\left(7-x\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\7-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}7-x< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x< 7\end{matrix}\right.\\\left\{{}\begin{matrix}x>7\\x< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3< x< 7\)

Vậy các số t/m điều kiện trên đều được.

tik mik nha !!!

3.

a,

\(25-\left(30+x\right)=1-\left(27-8\right)\\ 25-30-x=1-27+8\\ \left(-5\right)-x=-18\\ -x=-13\\ x=13\)

4,

a,

\(\left(x-3\right)\left(7-x\right)=0\Rightarrow\left[{}\begin{matrix}x-3=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

b,

\(\left(x+1\right)\left(x^2-4\right)=0\\ \left(x+1\right)\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)

Câu 1:

=>x^2=64

=>x=8(loại) hoặc x=-8(nhận)

a) \(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x-3\right)^3=\left(-2\right)^3\)

\(\Rightarrow x-3=-2\Leftrightarrow x=1\) vậy \(x=1\)

b) \(64x^3+48x^2+12x+1=0\Leftrightarrow\left(4x+1\right)^3=3^3\)

\(\Rightarrow4x+1=3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

d) \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\) vậy \(x=-2\)

e) \(x^3-6x^2+12x-8=0\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\) vậy \(x=2\)

bài2

a, x-15=-63-4

=>x-15=-67

=>x=-52

b, -x+3=11

=>x=-11+3

=>x=-8

c,\(|\)x+2\(|\)-4=7

=>\(|\)x+2\(|\)=11

=>\(\left\{{}\begin{matrix}x+2=11\\x+2=-11\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=9\\x=13\end{matrix}\right.\)

bài3

ta có:\(\left|y\right|\)=8

=>\(\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\)

TH1 x=5,y=8

=>x-y=5-8=-3

y-x=8-5=3

TH2x=5 ,y=-8

x-y=5--8=13

y-x=-8-5=-13

Baif:

a) x-15=-63-4

x-15=-67

x=-67+15

x=-52

b)-x+3=11

-x=11-3

-x=8

=> x=8

c)\(\left|x+2\right|-4=7\)

\(\left|x+2\right|\)=7+4=11

=> x+2=11 hoặc x+2=-11

x=11-2=9 hoặc x=-11-2=-13

Bài 3:

TH1: Nếu x=5 và y=8

thì x-y=5-8=-3

y-x=8-5=3

TH

: Nếu x=5 và y=-8

thì x-y=5-(-8)=13

y-x=(-8)-5=-13

a) x-15 = -63-4

x = -63-4+15

x = -52

b) -x + 3 = 11

-x= 11-3

-x= 8

x=8

c) | x+2 | -4 = 7

|x+2| = 11

⇒ \(\left\{{}\begin{matrix}x+2=11\\x+2=-11\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)