1: 7-x=8+(-7)

=>7-x=8-7=1

=>x=7-1=6

2: \(x-8=\left(-3\right)-8\)

=>x-8=-11

=>\(x=-11+8=-3\)

3: \(2-x=10-9+23\)

=>\(2-x=33-9=24\)

=>x=2-24=-22

4: \(-2-x=15\)

=>\(x=-2-15=-17\)

5: \(-7+x-8=-3-1+13\)

=>x-14=13-4=9

=>x=9+14=23

6: 100-x+7=-x+3

=>107-x=3-x

=>107=3(vô lý)

7: \(23+x=8-2x\)

=>\(x+2x=8-23\)

=>3x=-15

=>x=-15/3=-5

a: =1250:25=50

b: =234(45+55)+234=23400+234=23634

c: =1000x1000=1000000

97 chia x dư 7 => 90 chia hết cho x

Vậy x là ước chung của 234 và 90 và ước này phải lớn hơn 7

234 = 2.3².13

90 =  2.3².5

ƯC(234; 90) = {1; 2; 3; 6; 9; 18)

Vậy x = 9 hoặc x = 18

a: =234x(45+54+1)

=234x100

=23400

b: =(8x125)x(50x20)

=1000x1000

=1000000

a = 243x(45+54+1)

=243 x 100

=24300

b = (20 x 50 ) x (8x125)

=1000x1000

=1000000

a) 234 x 45 + 54 x 234 + 234

= 234 x (45+54+1) = 234 x 100 = 23400

b) 50 x 8 x 125 x 20 = (50 x 20) x (8 x 125)

= 1000 x 1000 = 1000000

a) Gọi x²=a 

=> 3a² - a - 234=0

∆=b² - 4ac= (-1)²-4×3×(-234)=2809

√∆=53

∆>0 nên pt có 2 nghiệm phân biệt 

a1=-b+√∆/2a = -(-1)+53/2×3 =9

a2=-b-√∆/2a = -(-1)-53/2×3 =-26/3

Thay x²=a=9 =>x=3,x=-3

          x²=a=-26/3 (loại)

Vậy nghiệm của pt là x =3, x=-3

d) (x+4)(x+5)(x+7)(x+8)=4

<=> (x+4)(x+8)(x+5)(x+7)=4

<=> (x²+8x+4x+32)(x²+7x+5x+35)=4

<=> (x²+12x+32)(x²+12x+35)=4

Đặt t=x²+12x+32

=> t(t+3)=4

<=> t²+3t-4=0

     (a=1,b=3,c=-4)

a+b+c=1+3+(-4)=0

=> t1=1 ; t2= c/a =-4/1=-4

Thay t=x²+12x+32=1

=> x²+12x+31=0

∆=b²-4ac= 12² -4×1×31= 20

√∆=2√5

∆>0 nên pt có 2 nghiệm phân biệt 

x1=-b+√∆/2a= -12+2√5/2×1= -6+√5

x2=-b-√∆/2a = -12-2√5/2×1= -6-√5

Thay t=x²+12x+32=-4

=> x²+12x+36=0

∆=b²-4ac= 12²-4×1×36=0

∆=0 nên pt có nghiệm kép 

x1=x2= -b/2a= -12/2×1 = -6

Vậy nghiệm của pt là S={-6+√5 ; -6-√5; -6}

b: =>\(\dfrac{-x}{x-1}=\dfrac{2x+3}{\left(x-1\right)\left(x+2\right)}\)

=>-x^2-2x-2x-3=0

=>x^2+4x+3=0

=>x=-1 hoặc x=-3

c: =>x^3+x^2-3x-3x^2+2x+2=0

=>x^3-2x^2-x+2=0

=>(x-2)(x-1)(x+1)=0

=>\(x\in\left\{2;1;-1\right\}\)

d: =>(x^2+12x+32)(x^2+12x+35)-4=0

=>(x^2+12x)^2+67(x^2+12x)+1116=0

=>(x^2+12x+36)(x^2+12x+31)=0

=>\(x\in\left\{-6;-6+\sqrt{5};-6-\sqrt{5}\right\}\)

a, |x-7|=|9-234|

=> |x-7|=|-225|

=> |x-7|=225

=>\(\orbr{\begin{cases}x-7=225\\x-7=-225\end{cases}}\)=>\(\orbr{\begin{cases}x=232\\x=-218\end{cases}}\)

b, \(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

=>\(\left(\frac{x}{8}\right)^2=\left(\frac{1}{2}\right)^2\)

=>\(\orbr{\begin{cases}\frac{x}{8}=\frac{1}{2}\\\frac{x}{8}=\frac{-1}{2}\end{cases}}\)=>\(\orbr{\begin{cases}x.2=8\\x.2=-8\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

c, (8-3²)²+(x-39)²=10

=>(-1)²+(x-39)²=10

=>(x-39)²=10-1

=>(x-39)²=9

=>(x-39)²=3²

=>\(\orbr{\begin{cases}x-39=3\\x-39=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=42\\x=36\end{cases}}\)

a)\(\left|x-7\right|=\left|9-234\right|\)

\(\Rightarrow\left|x-7\right|=\left|-225\right|\)

\(\Rightarrow\left|x-7\right|=225\)

\(\Rightarrow x-7=\pm225\)

Vậy \(x\in\left\{232;-218\right\}\)

b)\(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{4}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\left(\pm\frac{1}{2}\right)^2\)

\(\Rightarrow\frac{x}{8}=\pm\frac{1}{2}\)

Vậy \(x\in\left\{\pm4\right\}\)

c)\(\left(8-3^2\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(8-9\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(-1\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow1+\left(x-39\right)^2=10\)

\(\Rightarrow\left(x-39\right)^2=9\)

\(\Rightarrow\left(x-39\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow x-39=\pm3\)

Vậy \(x\in\left\{42;36\right\}\)

Lời giải:
1. Bạn coi lại đề

2. $=234\times (1257-257)=234\times 1000=234000$

3. $=175\times (1274-273-1)=175\times 1000=175000$

4. $=(4\times 25)\times (125\times 8)=100\times 1000=100000$

1. ...............................

2. $=234\times (1257-257)=234\times 1000=234000$

3. $=175\times (1274-273-1)=175\times 1000=175000$

4. $=(4\times 25)\times (125\times 8)=100\times 1000=100000$

a,234.4+234.5+234=234(4+5+1)=234.10=2340

b,135.16-135.2-135.4=135(16-4-2)=135.10=1350

c,1/2+1/4+1/8+1/16+1/32+1/64=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32=1-1/32=31/32

a) 234 x4 + 234 x 5 + 234 = 234 x ( 1+4+5)=2340

b)135 x16 -135 x2 -4 x135 =135 x (16-2-4) =1350

a) 234.(4+5+1)=234.10=2340

b)135.(16-2-4)=135.10=1350

\(a,234-\left(x-56\right)=789\)

\(\Leftrightarrow x-56=234-789\)

\(\Leftrightarrow x-56=-555\)

\(\Leftrightarrow x=\left(-555\right)+56=-499\)

Vậy x = -499

b) \(\frac{x+3}{-5}=\frac{x-15}{4}\)

\(\Leftrightarrow4\left(x+3\right)=-5\left(x-15\right)\)

\(\Leftrightarrow4x+12=-5x+75\)

\(\Leftrightarrow4x+12-\left(-5x\right)=75\)

\(\Leftrightarrow4x-\left(-5x\right)+12=75\)

\(\Leftrightarrow4x+5x=63\)

\(\Leftrightarrow9x=63\)

\(\Leftrightarrow x=7\)

Vậy x = 7

c) \(8\left(x-1\right)-7=2\left(x+2\right)+5\)

\(\Leftrightarrow8x-8-7=2x+4+5\)

\(\Leftrightarrow8x-8-7-2x+4=5\)

\(\Leftrightarrow8x-2x-8-7+4=5\)

\(\Leftrightarrow8x-2x=5-4+7+8\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

Vậy x = 4

d) Đặt \(D=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=\frac{2\left(x-1\right)+5}{x-1}=2+\frac{5}{x-1}\)

=> \(5⋮x-1\)

=> \(x-1\inƯ\left(5\right)\)

=> \(x-1\in\left\{\pm1;\pm5\right\}\)

=> \(x\in\left\{2;0;6;-4\right\}\)

D