`f)1-2m+m^2-x^2-4x-4`

`=(m^2-2m+1)-(x^2+4x+4)`

`=(m-1)^2-(x+2)^2`

`=(m-1-x-2)(m-1+x+2)`

`=(m-x-3)(m+x+1)`

`g)100a^2-(a^2+25)^2`

`=(10a)^2-(a^2+25)^2`

`=(-a^2+10a-25)(a^2+10a+25)`

`=-(a^2-10a+25)(a+5)^2`

`=-(a-5)^2+(a-5)^2`

`h)10x^2-7x-3`

`=(10x^2-10x)+(3x-3)`

`=10x(x-1)+3(x-1)`

`=(x-1)(10x+3)`

`j)(x^2+x)^2+4x^2+4x-12`

`=(x^2+x)^2+4(x^2+x)-12`

`=[(x^2+x)^2-2(x^2+x)]+[6(x^2+x)-12]`

`=(x^2+x)(x^2+x-2)+6(x^2+x-2)`

`=(x^2+x-2)(x^2+x+6)`

`=(x-1)(x+2)(x^2+x+6)`

`k)(x^2+x+1)(x^2+x+2)-12`

`=(x^2+x)^2+3(x^2+x)+2-12`

`=(x^2+x)^2+3(x^2+x)-10`

`=[(x^2+x)^2-2(x^2+x)]+[5(x^2+x)-10]`

`=(x^2+x)(x^2+x-2)+5(x^2+x-2)`

`=(x^2+x-2)(x^2+x+5)`

`=(x-1)(x+2)(x^2+x+5)` 

`a)20x^3y^4-5x^2y^3+100x^2y^4`

`=5x^2y^3(4xy-1+20y)`

`b)x^2-xy-15x+15y`

`=x(x-y)-15(x-y)`

`=(x-15)(x-y)`

`c)2x^3+8x^2+8x`

`=2x(x^2-4x+4)`

`=2x(x-2)^2`

`d)x^3-27y^6`

`=x^3-(3y^2)^3`

`=(x-3y)(x^2+3xy^2+9y^4)` 

`e)36x^2-a^2+10a-25`

`=36x^2-(a^2-10a+25)`

`=(6x)^2-(a-5)^2`

`=(6x-a+5)(6x+a-5)` 

`1)-3/5+7/10`

`=-6/10+7/10`

`=1/10`

`2)-7/12-8/15`

`=-35/60-32/60`

`=-67/60`

`3)0,75+1/8`

`=3/4+1/8`

`=6/8+1/8`

`=7/8`

`4)-3 1/8-(-0,12)`

`=-3-1/8+3/25`

`=-25/8+3/25`

`=-601/200`

`5)1,25+(-5/8)+(-5/2)`

`=5/4-5/8-5/2`

`=10/8-5/8-20/8`

`=-15/8`

`(2^43+2^4):(2^39+1)`

`=(2^4*2^39+2^4*1):(2^39+1)`

`=2^4*(2^39+1):(2^39+1)`

`=2^4`

`=16` 

Ta có: 

`(x+4)/(x-2)+(2x+5)/(x-2)` (x khác 2)

`=(x+4+2x+5)/(x-2)`

`=(3x+9)/(x-2)`

`=(3x-6+15)/(x-2)`

`=(3(x-2)+15)/(x-2)`

`=3+15/(x-2)`

Để giá trị của bt nguyên thì:

x - 2 ∈ Ư(15) = {1; -1; 3; -3; 5; -5; 15; -15}

=> x ∈ {3; 1; 5; -1; 7; -3; 17; -13} 

a) Đặt: `1/x=a;1/y=b` ta có:

\(\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\3\left(b+1\right)+4b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=b+1\\7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\b=\dfrac{2}{7}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{2}{7}+1=\dfrac{9}{7}\\b=\dfrac{2}{7}\end{matrix}\right.=>\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y}=\dfrac{2}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}\end{matrix}\right.\)

b) Đặt: `1/x=a;1/y=b` ta có:

\(\left\{{}\begin{matrix}6a+5b=3\\9a-10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12a+10b=6\\9a-10b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}21a=7\\6a+5b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\2+5b=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{5}\end{matrix}\right.=>\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)

`a)73^2-13^2-10^2+20*13`

`=73^2-(13^2+10^2-20*13)`

`=73^2-(13^2-2*13*10+10^2)`

`=73^2-(13-10)^2`

`=73^2-3^2`

`=(73+3)(73-3)`

`=76*70`

`=5320`

`b)x^2+8x+7`

`=(x^2+7x)+(x+7)`

`=x(x+7)+(x+7)`

`=(x+7)(x+1)`

Thay `x=49` vào bt ta có:

`(49+7)(49+1)=56*50=2800`

`78 xx 35 + 78 xx 23 + 58 xx 22`

`=78 xx (35 + 23) + 58 xx 22`

`=78 xx 58 + 58 xx 22`

`=58 xx (78 + 22)`

`=58 xx 100`

`=5800`

`9)4^16:16`

`=4^16:4^2`

`=4^14`

`10)6^12:36`

`=6^12:6^2`

`=6^10`

`11)7^8:49`

`=7^8:7^2`

`=7^6`

`12)8^4:64`

`=8^4:8^2`

`=8^2`

`13)9^7:81`

`=9^7:9^2`

`=9^5`

`14)2^40:32`

`=2^40:2^5`

`=2^35`

`15)3^100:27`

`=3^100:3^3`

`=3^97`

`16)5^25:25`

`=5^25:5^2`

`=5^23`

`17)8^19:64`

`=8^19:8^2`

`=8^17`

`18)12^8:144`

`=12^8:12^2`

`=12^6`

`19)121:11^2`

`=11^2:11^2`

`=1`

`20)10^100:1000`

`=10^100:10^3`

`=10^97`

`21)64*8^64`

`=8^2*8^64`

`=8^66`

`22)125*5^9`

`=5^3*5^9`

`=5^12`

`23)36*6^12`

`=6^2*6^12`

`=6^14`

`24)27*3^27`

`=3^3*3^27`

`=3^30` 

Ta có:

\(\left\{{}\begin{matrix}b\perp c\\a\perp c\end{matrix}\right.\)

`=>`a//b 

\(\Rightarrow\widehat{A}+\widehat{B}=180^o\) (cặp góc trong cùng phía) 

\(\Rightarrow m^0+4m^0=180^0\\ \Rightarrow5m^0=180^0\\ \Rightarrow m^0=\dfrac{180^0}{5}\\ \Rightarrow m^0=36^0\)

Chọn câu số 1