4) 121

5) - 32

6) 48

7) - 137

8) - 280

9) - 36

10) 42

4) 41

5) -32

6)  48

7) -137

8) -280

9) -36

10) 42

mấy bạn ghi cách viết giùm mình

cảm ơn

4) 87+(-45)-(-79)=  87 + ( -45 ) + 79 = 121

5) (-15)+45-62=  30 - 62 = -32

6) 120: (-5)+72=  -24 + 72 = 48

7) -147-130: (-13)= -147 - ( -10 ) = -137

8) 125. (-8) : (-25). (-7)= -1000 : (-25) (-7)= -280

9) 25-18+(-43) = 7 + ( -43) = -36

10) -18-3.(-20)= -18-( -60) = 42

đây lớp 6 sao

S

S

S

S

S

S

đề sai 

S

Đặt x+1=a rồi thay vào

\(\left(x+3\right)^4+\left(x-1\right)^4=626\)

Đặt \(a=x+1\)

\(\Rightarrow\)\(x+3=a+2\)

\(\Rightarrow\)\(x-1=a-2\)

ta có phương trình : 

\(\left(a-2\right)^4+\left(a+2\right)^4-626=0\)

Tới đây rồi thì dễ tự giải phương trình tiếp

=987

=4000

=76

1987-(538+462)=1987-1000=987

(2634-3634)+(873-4873)+(8374+626)=(-1000)+(-4000)+9000=4000

(2+18)+(4+16)+(6+10)+(12+8)=20+20+16+20=76

Sửa đề: \(\left(4,5-x\right)^4+\left(5,5-x\right)^4=1\)

=>\(\left(x-4,5\right)^4+\left(x-5,5\right)^4=1\)

=>\(\left\lbrack\left(x-5\right)+0,5\right\rbrack^4+\left\lbrack\left(x-5\right)-0,5\right\rbrack^4=1\) (1)

Đặt a=x-5; b=0,5

\(\left(a+b\right)^4+\left(a-b\right)^4\)

\(=\left(a^2+2ab+b^2\right)^2+\left(a^2-2ab+b^2\right)^2\)

\(=\left\lbrack\left(a^2+b^2\right)^2+2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack+\left\lbrack\left(a^2+b^2\right)^2-2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack\)

\(=2\left(a^4+2a^2b^2+b^4\right)+8a^2b^2=2a^4+12a^2b^2+2b^4\)

\(=2\left(x-5\right)^4+12\cdot\left(x-5\right)^2\cdot\left(0,5\right)^2+2\cdot\left(0,5\right)^4\)

\(=2\left(x-5\right)^4+3\left(x-5\right)^2+0,125\)

(1) sẽ trở thành: \(2\left(x-5\right)^4+3\left(x-5\right)^2+0,125=1\)

=>\(2\left(x-5\right)^4+3\left(x-5\right)^2-0,875=0\)

=>\(16\left(x-5\right)^4+24\left(x-5\right)^2-7=0\)

=>\(16\left(x-5\right)^4+28\left(x-5\right)^2-4\left(x-5\right)^2-7=0\)

=>\(\left\lbrack4\left(x-5\right)^2+7\right\rbrack\left\lbrack4\left(x-5\right)^2-1\right\rbrack=0\)

=>\(4\left(x-5\right)^2-1=0\)

=>\(\left(2x-10\right)^2=1\)

=>\(\left[\begin{array}{l}2x-10=1\\ 2x-10=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=11\\ 2x=9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{2}\\ x=\frac92\end{array}\right.\)

1,5+1-0,75