Tính hợp lí:

a, 0,75 . 1300 + 50 . 6,5-39

= 13 x (0,75 x 100) + 25 x (2 x 6,5) - 39

= 13 x 75 + 25 x 13 - 39

= 13 x 100 - 39

= 1300 - 39

= 1261

b, 8,4 . 84,5 + 840 . 0,15

b, 8,4 . 84,5 + 840 . 0,15

= 8,4 x 84,5 + 8,4 x (100 x 0,15)

= 8,4 x 84,5 + 8,4 x 15

= 8,4 x (84,5 + 15)

= 8,4 x 99,5

= 8,4 x 100 - 8,4 : 2

= 840 - 4,2

= 835,8

Bài 1:Phân tích các đa thức sau:

\(a,4x^2-6x\\ =2x(2x-3)\\ b,x^3y-2x^2y+5xy\\ = xy(x^2-2x+5)\\ c,2x^2(x+1) +4x(x+1)\\ =2x(x+1)(x+2)\\ d,\frac{2}{5}x.(y-1) -\frac{2}{5}x.(1-y)\\ =\frac{2}{5}x.(y-1)+\frac{2}{5}x.(y-1)\\ =2.\bigg[\frac{2}{5}x.(y-1)\bigg]\)

Bài 2 tính bằng cách hợp lý

\(a, 8,4.84, 5+840.0, 155\\ =8,4.(84,5+100.0,155)\\ =8,4.100\\ =840\\ b, 0,78.1300+50.6, 5-39\\ =(0,78.1300-39)+50. 6,5\\ =0,78.(1300-50)+50. 6,5\\ =0,78.1250+50. 6,5\\ =50.(0,78.25+6,5)\\ =1300\\ c,0, 12.90-110.0, 6+36-25.6\\ =6.(15.0,12-110.0,1+6-25)\\ =6.-28,2\\ =-169.2\)

Bài 3 Phân tích các đa thức sau\(a, (3x+1) ^2-(3x-1) ^2\\ =(3x+1-3x+1)(3x+1+3x-1)\\ =2.6x\\ b, (x+y) ^2-(x-y) ^2\\ =(x+y-x+y)(x+y+x-y)\\ =2y.2x\\ =2.(x-y)\\ c,(x+y)^3-(x-y) ^3\\ =(x+y-x+y)\bigg[(x+y)^2+(x+y)(x-y)+(x-y)^2\bigg]\\ =2y(x^2+2xy+y^2+x^2-xy+xy-y^2+x^2-2xy+y^2)\\ =2y(3x^2+y^2)\)

Bài 1:

\(a,4x^2-6x=2x\left(2x-3\right)\\ b,x^3y-2x^2y+5xy=xy\left(x^2-2x+5\right)\\ c,2x^2\left(x+1\right)+4x\left(x+1\right)=2x\left(x+1\right)\left(x+2\right)\\ d,\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(1-y\right)\\ =\frac{2}{5}\left(y-1\right)\left(x+y\right)\)

Bài 2:

\(a,8,4\cdot84,5+840\cdot0,155\\ =840\left(0,845+0,155\right)\\ =840\cdot1=840\\ b,0,78\cdot1300+50\cdot6,5-39\\ =39\cdot2\cdot13-39+25\cdot2\cdot6,5\\ =39\left(26-1\right)+25\cdot13\\ =39\cdot25+25\cdot13\\ =25\left(39+13\right)\\ =25\cdot52\\ =1300\\ c,0,12\cdot90-110\cdot0,6+36-25\cdot6\\ =6\cdot2\cdot0,9-6\cdot11+6\cdot6-25\cdot6\\ =6\left(1,8-11+6-25\right)\\ =-28,2\cdot6\\ =-169,2\)

Bài 3:

\(a,\left(3x+1\right)^2-\left(3x-1\right)^2\\ =\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\\ =2\cdot6x\\ =12x\\ b,\left(x+y\right)^2-\left(x-y\right)^2\\ =\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x\\ =4xy\\ c,\left(x+y\right)^3-\left(x-y\right)^3\\ =\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\)

a)      \({x^3} + 9{x^2} + 27x + 27 = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {\left( {x + 3} \right)^3}\)

Thay x=7 vào biểu thức ta được: \({\left( {7 + 3} \right)^3} = {10^3} = 1000\).

b)      \(27 - 54x + 36{x^2} - 8{x^3} = {3^3} - {3.3^2}.2x + 3.3.{\left( {2x} \right)^2} - {\left( {2x} \right)^3} = {\left( {3 - 2x} \right)^3}\)

Thay x=6,5 vào biểu thức ta được: \({\left( {3 - 2.6,5} \right)^3} = {\left( { - 10} \right)^3} =  - 1000\).

A = x² + 2x + 1 - y²

= (x + 1)² - y²

= (x + 1 - y) (x + 1 + y)

Với x = 6,5, y = 2,5, ta có:

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

= 5 . 10 = 50

Vậy khi x = 6,5, y = 2,5 thì A = 50.

bạn tham khảo nha

a)  85 .12,7 + 5.3.12,7 = 12,7 . ( 85 + 5.3 ) = 12,7 . (85+15) = 12,7 . 100 = 1270

b)  52.143 - 52.39 - 8.26 = 52.143 - 52.39 - 4.2.26 = 52.143 - 52.39 - 4.52 = 52.( 143 -39 - 4 ) = 52 . 100 = 5200

phân tích :

5x - 20y = 5 .(x - 4y)

5x(x-1)-3x.(x-1) = (x-1).(5x-3x) = 2x(x-1)  hoặc 5x(x-1)-3x(x-1) = x(x-1)(5-3) = 2x(x-1)

x.(x+y)-5x-5y = x(x+y) - 5 (x+y) = (x+y)(x-5)

a: \(x^2-3x=0\)

=>x(x-3)=0

=>x=0(nhận) hoặc x=3(nhận)

Thay x=0 vào A, ta được:

\(A=\frac{0-5}{0-4}=\frac{-5}{-4}=\frac54\)

Thay x=3 vào A, ta được:

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

b: \(B=\frac{x+5}{2x}-\frac{x-6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}\)

\(=\frac{x+5}{2x}+\frac{x-6}{x-5}-\frac{2x^2-2x-50}{2x\left(x-5\right)}\)

\(=\frac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\frac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}=\frac{x^2-10x+25}{2x\left(x-5\right)}\)

\(=\frac{\left(x-5\right)^2}{2x\left(x-5\right)}=\frac{x-5}{2x}\)

c: P=A:B

\(=\frac{x-5}{x-4}:\frac{x-5}{2x}=\frac{2x}{x-4}\)

Để P là số nguyên thì 2x⋮x-4

=>2x-8+8⋮x-4

=>8⋮x-4

=>x-4∈{1;-1;2;-2;4;-4;8;-8}

=>x∈{5;3;6;2;8;0;12;-4}

Kết hợp ĐKXĐ, ta được: x∈{3;6;2;8;12;-4}

\(D=50^2-49.51\)

\(\Leftrightarrow D=50^2-\left(50-1\right)\left(50+1\right)\)

\(\Leftrightarrow D=50^2-50^2+1=1\)

\(C=39^2+78.61+61^2\)

\(\Leftrightarrow C=39^2+2.39.61+61^2\)

\(\Leftrightarrow C=\left(39+61\right)^2=100^2=10000\)