`4)x^2-5x+6`

`=x^2-2x-3x+6`

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

`5)x^2+7x+10`

`=x^2+5x+2x+10`

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

`6)x+7\sqrt{x}+10`    `ĐK: x >= 0`

`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`

`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`

`7)3x^4+7x^2+4`

`=3x^4+3x^2+4x^2+4`

`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`

`8)x^2-x-2`

`=x^2-2x+x-2`

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

`9)x^6-x^3-2`

`=x^6+x^3-2x^3-2`

`=x^3(x^3+1)-2(x^3+1)`

`=(x^3+1)(x^3-2)`.

Phân tích đa thức thành nhân tử:

10)   \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)

\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)   (vì  1/86 + 1/85 + 1/84 + 1/83 + 1/4  \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy....

a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)

\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)

\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)

\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)

\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)

Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)

nên x-128=0

hay x=128

Vậy: x=128

b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)

\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)

\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)

Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)

nên x-1999=0

hay x=1999

Vậy: x=1999

a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4

<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)

<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)

<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)

<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)

Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128

b)Tương tự

<=>x-128=0

<=>x=128

Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0

b)tương tự

a =(3427+343)-1247

   =3770          - 2523

   =       1247

B , =(5327-247)-(1247-327)

      =    5080     -  920

       = kết quả tự đi mà tính

C ) = tự đi mà làm người ta giúp 2 câu ròi

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)            (vì 1/75 + 1/70 - 1/65 - 1/60  \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy.....

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)    (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy...

9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)

=>x-99=0

hay x=99

7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)

=>x+100=0

hay x=-100

8:

Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\) 

\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)

=>200-x=0

hay x=200

a. 523 – x = 301

x = 523 – 301

x = 222

b. x : 642 = 6

x = 642 x 6

x = 3 852

476 x 476 + 523 x 476 + 476 = 476 x ( 476 + 523 + 1 ) = 476 x 1000 = 476000

\(=476\times476+523\times476+1\times476\)

\(=\left(476+523+1\right)\times476\)

\(=1000\times476\)

\(=476000\)

= 476 x 476 + 523 x 476 + 476 x 1

= 476 x (523 + 476 + 1 )

= 476 x 1000

= 476000