2: 

a: =>x=1/2+1/6=3/6+1/6=4/6=2/3

b: =>x+3,5=7,8-6,3=1,5

=>x=1,5-3,5=-2

c: =>(35-x)/-105=1/5

=>35-x=-21

=>x=56

1

c hình như tính bình thường thôi:v

d

= 1,75 : 5 + 2,5 . (16 – 4 . 4,1)

= 1,75 : 5 + 2,5 . (16 – 16,4)

= 1,75 : 5 + 2,5 . (−0,4)

= 0,35 − 1

= −0,65.

2

a

\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

b

\(\Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{5}:2\\ \Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{10}=\dfrac{15}{10}\\ \Rightarrow x=\dfrac{15}{10}-\dfrac{35}{10}=-2\)

c không rõ đề:v

\(1,\dfrac{2}{-11}.6\dfrac{2}{7}+\dfrac{3}{7}:3+5\\ =\dfrac{2}{-11}.\dfrac{44}{7}+\dfrac{3}{7}\times\dfrac{1}{3}+5\\ =-\dfrac{88}{77}+\dfrac{3}{21}+5\\ =-\dfrac{8}{7}+\dfrac{1}{7}+5\\ =-1+5\\ =4\)

\(d,1,75:5+2,5.\left(4^2-4.4,1\right)\\ =0,35+2,5.\left(16-16,4\right)\\ =0,35+2,5.0,4\\ =0,35+1\\ =1,35\)

`2,`

`a, 1/2 -x=-1/6`

`=> x= 1/2 -(-1/6)`

`=> x= 1/2 +1/6`

`=> x= 3/6+1/6`

`=>x= 4/6`

`=>x=2/3`

Vậy `x=2/3`

`b, x+3,5 =7,8 -12,6 :2`

`=> x+3,5 = 7,8 - 6,3`

`=> x+3,5=1,5`

`=> x= 1,5-3,5`

`=>x=-2`

Vậy `x=-2`

\(c,\dfrac{7}{35}=\dfrac{35-x}{-105}\\ \Rightarrow7.\left(-105\right)=35.\left(35-x\right)\\ \Rightarrow-735=1225-35x\\ \Rightarrow-735-1225=-35x\\ \Rightarrow-1960=-35x\\ \Rightarrow x=-1960:\left(-35\right)\\ \Rightarrow x=56\)

Vậy `x=56`

Lần sau đăng từng bài thôi bạn nhé :)

Đề đúng đây chứ ?

\(3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)

\(=3\left[\left(x^2+y^2+2xy\right)-2xy\right]-2\left[\left(x^3+3x^2y+3xy^2+y^3\right)-3xy\left(x+y\right)\right]\)

\(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\)

Thay \(x+y=1;\)có :

\(=3\left(1-2xy\right)-2\left(1-3xy\right)\)

\(=3-6xy-2+6xy=3-2=1\)

Vậy ...

\(\frac{\left(a-b+c\right)^2}{4}+\frac{\left(a+b-c\right)^2}{4}+\frac{\left(-a+b+c\right)^2}{4}\)

\(=\frac{\left(a+b+c-2b\right)^2}{4}+\frac{\left(a+b+c-2c\right)^2}{4}+\frac{\left(a+b+c-2a\right)^2}{4}\)

\(=\frac{\left(4m-2b\right)^2}{4}+\frac{\left(4m-2c\right)^2}{4}+\frac{\left(4m-2a\right)^2}{4}\)

\(=\frac{16m^2+4b^2-16bm}{4}+\frac{16m^2+4c^2-16cm}{4}+\frac{16m^2+4a^2-16am}{4}\)

\(=4m^2+b^2-4bm+4m^2+c^2-4cm+4m^2+a^2-4am\)

\(=12m^2+b^2+c^2+a^2-4m\left(a+b+c\right)\)

\(=12m^2+b^2+c^2+a^2-4m\left(4m\right)\)

\(=a^2+b^2+c^2-4m^2\)

Chắc hết rồi nhỉ :/

2

a

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)

\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)

b

Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)

Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)

Áp dụng kết quả câu a ta được:

\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

AB=(3;-1)

AC=(4;2)

AB.AC= |AB|.|AC|.cos(AB,AC)

cos( AB,AC)= \(\dfrac{10}{\sqrt{10}.2\sqrt{5}}=\dfrac{\sqrt{2}}{2}\)

BAC=45