a.

$x^2+20x+100=(x+10)^2$

b.

$16x^2+24xy+9y^2=(4x+3y)^2$

c.

$y^2-14y+49=(y-7)^2$
d.

$9x^2-42xy+49y^2=(3x-7y)^2$

e.

$4x^2-9y^2=(2x-3y)(2x+3y)$

f.

$16-x^2=(4-x)(4+x)$

g.

$49x^2-1=(7x-1)(7x+1)$

h.

$16x^2-25=(4x-5)(4x+5)$

i.

$8x^3+24x^2y+54xy^2+27y^3=(2x+3y)^3$

k.

$x^3-6x^2y+12xy^2-8y^3=(x-2y)^3$

l.

$(2a+b)(4a^2-2ab+b^2)=(2a)^3+b^3=8a^3+b^3$

m.

$(3x-4y)(9x^2+12xy+16y^2)=(3x)^3-(4y)^3=27x^3-64y^3$

(3x-1)^7=(1-3x)

(3x-1)^7 -(1-3x)=0

(3x-1)^7+(3x-1)=0

(3x-1).[(3x-1)^6+1]=0

TH1: 3x-1=0                     TH2 :   (3x-1)^6+1=0

       3x    =1                               (3x-1)^6  =-1 suy ra x\(\in\) rỗng (bạn viết kí hiệu nhe xuống dòng nữa)

        x     =1:3=1/3

vậy x = \(\frac{1}{3}\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{16}=0\)

\(\Leftrightarrow\left(x-7\right)^{16}.\left(x-7\right)^{x-15}-\left(x-7\right)^{16}=0\)

\(\Leftrightarrow\left(x-7\right)^{16}\left[\left(x-7\right)^{x-15}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{16}=0\\\left(x-7\right)^{x-15}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{x-15}=1^{x-15};\left(x-7\right)^{x-15}=\left(x-7\right)^0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x-7=1;x-15=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=8;x=15\end{cases}}\)

Vậy \(x\in\left\{7;8;15\right\}\)

P/s: Thay cái ngoặc có 2 nhánh thành ngoặc 3 nhánh cho nó đẹp :))

\(2n+3⋮3n+4\Leftrightarrow6n+9⋮3n+4\)

\(\Leftrightarrow2\left(3n+4\right)+1⋮3n+4\Leftrightarrow1⋮3n+4\)

\(\Rightarrow3n+4\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(2n+3⋮3n+4\)

Ta có: \(2n+3=3\left(2n+3\right)=6n+9\)

\(3n+4⋮3n+4\Leftrightarrow2\left(3n+4\right)⋮3n+4\Leftrightarrow6n+8⋮3n+4\Leftrightarrow\left(6n+9\right)-\left(6n+8\right)⋮3n+4\)

\(\Leftrightarrow1⋮3n+4\Leftrightarrow3n+4\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow n\in\left\{-1;\frac{-5}{3}\right\}\)

Đầu bài thế này thì tìm x đến tết chả hết

`a)`

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

`=2x^{2}+x-1`

`=2(x^{2}+(1)/(2)x-(1)/(2))`

`=2(x^{2}+(1)/(2)x+(1)/(16)-(9)/(16))`

`=2(x+(1)/(4))^{2}-(9)/(8)>= -9/8` với mọi `x`

Dấu `=` xảy ra khi :

`x+(1)/(4)=0<=>x=-1/4`

Vậy `min=-9/8<=>x=-1/4`

``

`b)`

`(4x+1)(2x-5)`

`=8x^{2}-18x-5`

`=8(x^{2}-(9)/(4)x-(5)/(8))`

`=8(x^{2}-(9)/(4)x+(81)/(64)-(121)/(64))`

`=8(x-(9)/(8))^{2}-(121)/(8)>= -(121)/(8)` với mọi `x`

Dấu `=` xảy ra khi :

`x-(9)/(8)=0<=>x=9/8`

Vậy `min=-121/8<=>x=9/8`

\(A=2x^2+x-1=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(A_{min}=-\dfrac{9}{8}\) khi \(x=-\dfrac{1}{4}\)

\(B=8x^2-18x-5=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(B_{min}=-\dfrac{121}{8}\) khi \(x=\dfrac{9}{8}\)

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{9}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)

b) \(B=\left(4x+1\right)\left(2x-5\right)=8x^2-18x-5=8\left(x^2-\dfrac{9}{4}x+\dfrac{81}{64}\right)-\dfrac{121}{8}=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\)

Vì \(8\left(x-\dfrac{9}{8}\right)^2\ge0\Rightarrow8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{9}{8}\)

\(a,\Rightarrow x\times\left(42+54+4\right)=25400\\ \Rightarrow x\times100=25400\\ \Rightarrow x=254\\ b,\Rightarrow x\times\left(142-41-1\right)=40800\\ \Rightarrow x\times100=40800\\ \Rightarrow x=408\)

a) \(x\)(42+54+4)=15400            b) (142-41-1)\(x\)=10800
\(x\)x100=15400                             100x\(x\)=10800
\(x\)=15400:100                                      \(x\)=10800:100
\(x\)=154                                                 \(x\)=108