a) \(97\cdot13+130\cdot0,3\)

\(=97\cdot13+13\cdot3\)

\(=13\cdot\left(97+3\right)\)

\(=13\cdot100\)

\(=1300\)

b) \(86\cdot153-530\cdot8,6\)

\(=86\cdot153-53\cdot86\)

\(=86\cdot\left(153-53\right)\)

\(=86\cdot100\)

\(=8600\)

a) \(97.13+130.0,3\)

\(=97.13+13.3\)

\(=13\left(97+3\right)=13.100=1300\)

b) \(86.153-530.8,6\)

\(=86.153-53.86\)

\(=86.\left(153-53\right)=86.100=8600\)

a. 97 . 13 + 130  . 0,3 = 97 . 13 + 13 . 10 . 0,3

                                = 97 . 13 + 13 . 3

                                = 13 . ( 97 + 3 ) 

                                = 13 . 100

                                = 1300

b. 86 . 153 - 530 . 8,6  = 86 . 153 - 53 . 10 .8,6

                                 = 86 . 153 - 53 .86

                                 = 86 ( 153 - 53 )

                                 = 86 . 100

                                 = 8600

Bài làm

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

97 x 13 + 130 x 0,3

= 97 x 13 + 13 x 3

= 13 x (97 + 3)

= 13 x 100 = 1300

\(97\times13+130\times0.3\)

\(=97\times13+13\times10\times0.3\)

\(=97\times13+13\times30\)

\(=13\times\left(97+30\right)\)

\(=13\times127\)

\(=1651\)

a, Ta có : \(\frac{x+2}{89}+\frac{x+5}{86}=\frac{x+8}{83}+\frac{x+11}{80}\)

=> \(\frac{x+2}{89}+1+\frac{x+5}{86}+1=\frac{x+8}{83}+1+\frac{x+11}{80}+1\)

=> \(\frac{x+91}{89}+\frac{x+91}{86}=\frac{x+91}{83}+\frac{x+91}{80}\)

=> \(\frac{x+91}{89}+\frac{x+91}{86}-\frac{x+91}{83}-\frac{x+91}{80}=0\)

=> \(\left(x+91\right)\left(\frac{1}{89}+\frac{1}{86}-\frac{1}{83}-\frac{1}{80}\right)=0\)

=> \(x+91=0\)

=> \(x=-91\)

Vậy phương trình trên có nghiệm là \(S=\left\{-91\right\}\)

b, Ta có : \(\frac{x-11}{99}+\frac{x-12}{98}=\frac{x-3}{97}+\frac{x-4}{96}\)

=> \(\frac{x-11}{99}-1+\frac{x-12}{98}-1=\frac{x-3}{97}-1+\frac{x-4}{96}-1\)

=> \(\frac{x-110}{99}+\frac{x-110}{98}=\frac{x-110}{97}+\frac{x-110}{96}\)

=> \(\frac{x-110}{99}+\frac{x-110}{98}-\frac{x-110}{97}-\frac{x-110}{96}=0\)

=> \(\left(x-110\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

=> \(x-110=0\)

=> \(x=110\)

Vậy phương trình trên có nghiệm là \(S=\left\{110\right\}\)

rất cảm ơn bn

Cảm ơn bạn

Ta có: 

\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-\left(ab+bc+ca\right)+\left(a+b+c\right)-1\)

\(=abc-\left(ab+bc+ca\right)+5\ge0\)

\(\Rightarrow abc\ge ab+bc+ca-5\)(1)

\(\left(a-3\right)\left(b-3\right)\left(c-3\right)=abc-3\left(ab+bc+ca\right)+9\left(a+b+c\right)-27\)

\(=abc-3\left(ab+bc+ca\right)+27\le0\)

\(\Rightarrow abc\le3\left(ab+bc+ca\right)-27\)(2)

(1)(2) suy ra \(ab+bc+ca-5\le3\left(ab+bc+ca\right)-27\)

\(\Leftrightarrow ab+bc+ca\ge11\).

\(6^2=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2=36-2\left(ab+bc+ca\right)\le36-2.11=14\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\\\left(a-3\right)\left(b-3\right)\left(c-3\right)=0\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}}\)và các hoán vị. 

a=1

b=2

c=3