Ta có : x + 4 > x - 9 

\(\left\{{}\begin{matrix}x+4>0\\x-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-4\\x< 9\end{matrix}\right.\)<=> -4 < x < 9 

\(\left(x+4\right)\left(x-9\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4< 0\\x-9>0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-4\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< -4\\x>9\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-4< x< 9\)

\(\left(x-1\right)^{x+1}-\left(x-1\right)^{x+12}=0\\ \Leftrightarrow\left(x-1\right)^{x+1}\left[1-\left(x-1\right)^{11}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+11}=0\\\left(x-1\right)^{11}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

đây là toán lp 6 hk pải lp 3 dễ gây hiểu lầm nha bn

câu này dễ mà  bạn

re thi lam di chi gioi noi thoi

Pham minh quang : quá giỏi ns 

a)1/2:0,5-1/4:0,25+1/8:0,125-1/10:0,1

=1/2x2-1/4x4+1/8x8-1/10x10

=1-1+1-1

=0

b)không biết k 2 cái là dc rồi

a) (3x – 2)(4x + 5) = 0

 ⇔ 3x – 2 = 0 hoặc 4x + 5 = 0

 ⇔ 3x = 2 hoặc 4x  = -5

 ⇔ x = \(\dfrac{2}{3}\) hoặc x  = \(\dfrac{-5}{4}\)

 Vậy tập nghiệm là S = {\(\dfrac{2}{3}\); \(\dfrac{-5}{4}\)}

b) 2x(x – 3) + 5(x – 3) = 0

 ⇔ (x – 3)(2x + 5) = 0

 ⇔ x – 3 = 0 hoặc 2x + 5 = 0

 ⇔ x = 3 hoặc 2x = \(-5\)

 ⇔ x = 3 hoặc x = \(\dfrac{-5}{2}\)

 Vậy tập nghiệp là S = {3; \(\dfrac{-5}{2}\)}

\(2^{10}.13+2^{10}.65:2^8.104=2^{10}.78:2^8.104=2^2.78.104=4.78.104=32448\)

aVT=.\(\left(a+b+c\right)^2+a^2+b^2+c^2\)

=\(a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)

=\(2a^2+2b^2+2c^2+2ab+2ac+2bc\)

VP=\(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)=\(a^2+2ab+b^2+b^2+2bc+b^2+a^2+2ac+c^2\)

=\(2a^2+2b^2+2c^2+2ab+2bc+2ac\)

Vậy VT=VP

a)\(\text{(a+b+c)^2 +a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2}\)

Ta có:

\(\left(a+b+c\right)^2+a^2+b^2+c^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)

\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ca+a^2\right)\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

Vậy \(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

b) Câu b sao chỉ có một vế vậy , hằng đẳng thức thì phải có hai vế chứ

b) \(\text{x^4+y^4+(x+y)^4=2(x^2+xy+y^2)^2}\)

Ta có:

\(x^4+y^4+\left(x+y\right)^4=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

\(2x^4+2y^{\text{4}}+4x^3y+6x^2y^2+4xy^3=2\left(x^4+y^4+2x^3y+3x^2y^2+2xy^3\right)\)

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

\(=2\left(x^2+xy+y^2\right)^2\)

Vậy \(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)