a) 25-5x+3-4=0

=> 24-5x=0

=> 5x=24

=> x=\(\dfrac{24}{5}\)

b) 259-51/4x+3/=100.3+10=310

=> -51/4x+3/=310-259=51

=> /4x+3/=-51 (1)

Vi /4x+3/≥0 (2)

Tu (1)(2) => pt vo nghiem

c) 7x-21-15+5x=6

=> 12x-36-6=0

=> 12x-42=0

=> 12x=42

=> x=\(\dfrac{42}{12}=\dfrac{7}{2}\)

Bài 2: 

Ư(42)={1;2;3;6;7;14;21;42}

mà số cần tìm lớn hơn 10

nên A={14;21;42}

Bài 1:

\(\Leftrightarrow259-51\left|4x+3\right|=100\cdot3+10=310\)

=>51|4x+3|=259-310<0

=>\(x\in\varnothing\)

\(259-51\left|4x+3\right|=\left(-10\right).3+10\)

\(\Leftrightarrow259-51\left|4x+3\right|=-30+10\)

\(\Leftrightarrow259-51\left|4x+3\right|=-20\)

\(\Leftrightarrow-51\left|4x+3\right|=-20-259\)

\(\Leftrightarrow-51\left|4x+3\right|=-279\)

\(\Leftrightarrow\left|4x+3\right|=\frac{279}{51}\)

\(\Leftrightarrow\orbr{\begin{cases}4x+3=\frac{93}{17}\\4x+3=-\frac{93}{17}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{21}{34}\\x=-\frac{36}{17}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{21}{34}\\x=-\frac{36}{17}\end{cases}}\)

1.4./x-5/-54=-34

4./x-5/=-34+54

4./x-5/=20

/x-5/=20:4

/x-5/=5

=>x-5=5 va x-5=-5

TH1:x-5=-5

=>x=0

TH2:x-5=5

=>x=10

2. /x-1/+(-5)=2

/x-1/=2-(-5)

/x-1/=7

=>x-1=7 va x-1=-7

TH1: x-1=7

=>x=8

TH2:x-1=-7

=>x=-6

vay x=-6 hoac x=8

3. 32-8./2x-9/=24

8./2x-9/=32-24=8

/2x-9/=8:8=1

=>2x-9=1 hoac 2x-9=-1

TH1: 2x-9=1

=>x=(1+9):2=5

TH2: 2X-9=-1

=>x=(-1+9):2=4

vay x=4 hoac x=5

\(a,\)

\(4.\left|x-5\right|-54=-34\)

\(4.\left|x-5\right|\)        \(=-34+\left(-54\right)\)

\(4.\left|x-5\right|\)        \(=-88\)

    \(\left|x-5\right|\)        \(=-88\div4\)    

    \(\left|x-5\right|\)        \(=-22\)

\(\Rightarrow\orbr{\begin{cases}x-5=22\\x-5=-22\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=22+5\\x=-22+5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=27\\x=-17\end{cases}}\)

1) (-10)-|5-x|=(-12)

|5-x|=(-10)-(-12)

|5-x|=2

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

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

Vậy:............

mik nghĩ chắc trả ai rảnh để trả lời hết đâu dù biết làm

mình cũng nghỉ thế

me too

1. Thực hiện phép tính bằng cách hợp lí :

a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]

= 0+(-100) = -100

b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75

= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000

c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]

= (-151) + [(-100) + 100] = -151

d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]

= 32+[(-200)+(-400)] = 32+(-600) = -568

e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65

= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)

= 65.(-70) = -4550

g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43

= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)

= 16.(-10) = -160

\(1,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\\ 2,4x^2-4x+1-25y^2=\left(2x-1\right)^2-\left(5y\right)^2=\left(2x-5y-1\right)\left(2x+5y-1\right)\\ c,x^2-3x-10=\left(x^2-5x\right)+\left(2x-10\right)=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+2\right)\)