= -251 . 3 - 281 + 3 . 251 - 1 + 281

ấn nhầm

A= (3,1 - 2,5) - (-2,5 + 3,1)

=3.1-2.5+2.5-3.1

=0

B= (5,3 - 2,8) - (4 + 5,3)

=5.3 - 2.8-4-5.3

= -6.8

C= -(251.3 + 281) + 3.251 - (1-281)

=-251.3-281+3.251-1+281

=-1

D= -(3/5+3/4)−(−3/4+2/5)

= -3/5-3/4+3/4-2/5

= -5/5

=-1

\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)=3,1-2,5+2,5-3,1=0\)

\(B=\left(5,3-2,8\right)-\left(4+5,3\right)=5,3-2,8-4-5,3=-6,8\)

\(C=-\left(215\cdot3+281\right)+3\cdot215-\left(1-281\right)=-215\cdot3-281+3\cdot215-1+281=1\)

\(D=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)

Bài 1 : Tính giá trị của các biểu thức sau đã bỏ dấu ngoặc :

A= (3,1 - 2,5) - (- 2,5 + 3,1) = 3,1 - 2,5 + 2,5 - 3,1 = 0

B= (5,3 - 2,8) - (4 + 5,3) = 5,3 - 2,8 - 4 - 5,3 = - 6,8

C= - (251.3 + 281) + 3.251 - (1 - 281) = - 753 - 281 + 753 + 280 = - 1

D= - (35 + 34) − ( − 34 + 25) = - 35 - 34 + 34 - 25 = - 60

=-251.3-281+3.251-1+281

=(-251.3+3.251)+(281-281)-1

=0+0-1

=-1

\(C=\left(-251\cdot3+251\cdot3\right)+\left(-281+281\right)-1=-1\)

A = \(\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)

\(=3,1-2,5+2,5-3,1=6,2\)

B = \(\left(5,3-2,8\right)-\left(4+5,3\right)\)

\(=5,3-2,8-4-5,3=-6,8\)

C = \(-\left(251.3+281\right)+3.251-\left(1-281\right)\)

\(=-251.3-281+3.251-1+281=-1\)

D = \(-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)\)

\(=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)

(Nhớ k cho mình với nhá!)

a) (4,1 - 2,5 ) - ( -2,5 + 3,1 ) = 1,6 - 0,6 = 1

b) ( 5,3 - 2,8) - ( 4+ 5,3) = 2,5 - 9,3 = -6.8

c) -( 251,3 + 281) + 3,251 - (1 - 981) = -532,3 + 3,251 + 980 = 450,951

d) -(3/5 + 3/4) - ( -3/4 +2/5) = -3/5 - 3/4 + 3/4 - 2/5 = -1

Bài 1: 

\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\) 

\(=\frac{1}{\frac{1}{2}}+3\)  \(=2+3\) \(=5\)

                                                  Vậy B=5

Bài 2:

a) x³ - 36x = 0  

=>  x(x²-36)=0

=>  x(x²+6x-6x-36)=0 

=> x[x(x+6)-6(x+6) ]=0

=> x(x+6)(x-6)=0

\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)

 \(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)

                                  Vậy x=0; x=-6; x=6

b)  (x - y = 4 => x=4+y)

 x−3y−2 =32  

=>2(x-3) = 3(y-2)

=>2x-6= 3y-6

=>2x-3y=0

=>2(4+y)-3y=0

=>8+2y-3y=0

=>8-y=0

=>y=8 (thỏa mãn)

Do đó x=4+y=4+8=12 (thỏa mãn)

         Vậy x=12 và y =8

B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4  1/5 - 1/8 

B= 1/ 1/2 + 3

B= 2+3

B=5

B2:

a) x^3 - 36x = 0

x(x^2 - 36) = 0

=> x=0  hoặc x^2-36=0

=> x= 0 hoặc x^2=36

=> x=0 hoặc x= +- 6

b) x-y = 4 => x= 4+y

thay x=4+y vào x- 3/ y-2=3/2, có:

4+y-3/ y+2 = 3/2

y+1/ y+2 = 3/2

y+2 -1/ y+2 = 3/2

1 - 1/y+2 = 3/2

1/y+2= 1-3/2

1/y+2 = -1/2

=> y+2 = -2

=> y= -4

Dp x= 4+y => x= 4-4

=> x=0

Vậy x=0 và y=-4

\(a,\left(x+1\right)^2=81\) 

    \(\left(x+1\right)^2=9^2\)  Hoặc \(\left(x+1\right)^2=\left(-9\right)^2\)

      \(\left(x+1\right)=9\)                     \(x+1=-9\)

                     \(x=8\)                               \(x=-10\)

b,\(\left(x+5\right)^{^{ }3}=-64\)

  \(\left(x+5\right)^3=\left(-4\right)^3\)

          \(x+5=-4\)

=>               \(x=-9\)

c,\(\left(2x-3\right)^2=9\)

=>\(\left(2x-3\right)^2=3^2\)Hoặc  \(\left(2x-3\right)^2=\left(-3\right)^2\)

            \(2x-3=3\)                    \(2x-3=-3\)

                     \(2x=6\)                             \(2x=0\)       

=> \(\hept{\begin{cases}x=3\\x=0\end{cases}}\)

d, \(\left(4x+1\right)^3=27\)

   \(\left(4x+1\right)^{^{ }3}=3^3\)

            \(4x+1=3\)

                     \(4x=2\)

                       \(x=\frac{1}{2}\)

\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{8^6}{4}=\frac{\left(2^3\right)^6}{2^2}=\frac{2^{18}}{2^2}=2^{16}\)

\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{15}+4^{10}}{4^6+4^{11}}=\frac{4^{10}.4^5+4^{10}}{4^6+4^6.4^5}=\frac{4^{10}.\left(4^5+1\right)}{4^6.\left(4^5+1\right)}=\frac{4^{10}}{4^6}=4^4=256\)

phần D trên mk làm sai xin lỗi nha