mk ko bt 123

Ta có $\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=2$

$=>2M=(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15})(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15})$

$=>2M=\sqrt{x^2-7x+19}^2-\sqrt{x^2-7x+15}^2$

$=>2M=(x^2-7x+19)-(x^2-7x+15)=4$

$=>M=2$

\(2.M=\left(x^2-7x+19\right)-\left(x^2-7x+15\right)=4\Rightarrow M=2\)

Đặt \(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=B\)  = B

   Xét tích \(AB=\left(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}\right)\left(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15}\right)\)

                   \(=x^2-7x+19-\left(x^2-7x+15\right)=x^2-7x+19-x^2+7x-15\)

                     \(=4\)

     Mà \(B=2\Leftrightarrow A=2\)

Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

\(=x^2+4xy+4y^2\)

e) \(\left(x-3y\right)\left(x+3y\right)\)

\(=x^2-\left(3y\right)^2\)

\(=x^2-9y^2\)

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

\(=25-10x+x^2\)

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-6xy\right)\)

Bài `12`

`(1/2x+4)^2`

`=(1/2x)^2 + 2 . 1/2x.4 + 4^2`

`= 1/4 x^2 +4x + 16`

__

`(7x-5y)^2`

`=(7x)^2-2.7x.5y+(5y)^2`

`= 49x^2 -  70xy   + 25y^2`

__

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

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

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

`=y^4-36x^4`

__

`(x+2y)^2`

`=x^2+ 2.x.2y+(2y)^2`

`= x^2 + 4xy +4y^2`

__

`(x-3y)(x+3y)`

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

`=x^2 - 9y^2`

__

`(5-x)^2`

`=5^2 -2.5.x+x^2`

`=25 - 10x+x^2`

Bài `11`

`(7x+4)^2 -(7x+4)(7x-4)`

`= (7x+4)(7x+4) -(7x+4)(7x-4)`

`=(7x+4)(7x+4-7x+4)`

`=8(7x+4)`

`= 56x+32`

__

`(x+2y)^2-6xy (x+2y)`

`= (x+2y) (x+2y-6xy)`

thảo có ghi lộn đề k

\(\left(7x+3\right)^2-\left(7x-1\right)\left(7x-3\right)=-12\)

\(\Rightarrow49x^2+42x+9-\left(49x^2-21x-7x+3\right)=-12\)

\(\Rightarrow70x+18=0\) \(\Rightarrow x=-\dfrac{18}{70}=-\dfrac{9}{35}\)

Với x = 6 ta có

A= 6⁵ - 7.6⁴ + 7.6³ - 7.6² + 7.6 - 1

  = 6⁵ - (6+1).6⁴ + (6+1).6³ - (6+1).6² + (6+1).6 - 1

  = 6⁵ - 6⁵ - 6⁴ + 6⁴ + 6³ - 6³ - 6² + 6² + 6 - 1

  = 5

Tớ biết làm đúng 100%:

\((x\cdot1+x\cdot\frac{7}{9})\left(x\cdot1+x\cdot\frac{7}{20}\right)...\left(x\cdot1+x\cdot\frac{7}{9200}\right)=\frac{186}{25}\)

\(x\cdot\left(1+\frac{7}{9}\right)\cdot x\left(1+\frac{7}{20}\right)\cdot...\cdot x\left(1+\frac{7}{9200}\right)=\frac{186}{25}\)

\(\left(x\cdot x\cdot...\cdot x\right)(\frac{16}{9}+\frac{27}{20}+...+\frac{9207}{9200})=\frac{186}{25}\)

\(\left(x\cdot x\cdot...\cdot x\right)\left(\frac{2\cdot8}{1\cdot9}+\frac{3\cdot9}{2\cdot10}+...+\frac{93\cdot99}{92\cdot100}\right)=\frac{186}{25}\)

\(x^{92}\cdot\frac{2\cdot8\cdot3\cdot9\cdot...\cdot93\cdot99}{1\cdot9\cdot2\cdot10\cdot...\cdot92\cdot100}=\frac{186}{25}\)

\(x^{92}\cdot\frac{\left(2\cdot3\cdot...\cdot93\right)\cdot\left(8\cdot9\cdot...\cdot99\right)}{\left(1\cdot2\cdot...\cdot92\right)\cdot\left(9\cdot10\cdot...\cdot100\right)}=\frac{186}{25}\)

\(x^{92}\cdot\frac{93\cdot8}{100}=\frac{186}{25}\)

\(x^{92}\cdot\frac{186}{25}=\frac{186}{25}\)

\(x^{92}=\frac{186}{25}:\frac{186}{25}\)

\(x^{92}=1\Rightarrow x=1\)

cô tớ giải rồi . x=1 (đúng 100%)

CÂU NÀY X=1 ĐÓ !

CẬU KIA ĐÚNG RỒI !

7x.(2+x)-7x.(x+3)=14

7x.[(2+x)-(x+3)]=14

7x.[x+2-x-3]=14

7x.(-1)           =14

7x                 =14:(-1)

7x                 =-14

 x                    =-14:7

x                     =-2

Chúc bn học tốt

 \(7x\cdot\left(2+x\right)-7x\cdot\left(x+3\right)=14\)

                \(7x\cdot\left(2+x-x-3\right)=14\)

                                       \(7x\cdot\left(-1\right)=14\)

                                                        \(7x=-14\)

                                                           \(x=-2\)

Ta có : 7x. ( 2+x ) - 7x. ( x+3 ) = 14

     <=> 7x . \([(2+x)-\left(x+3\right)]\)=14

<=>7x . \([2+x-x-3]\)=14

<=> 7x. (-1) = 14

<=> 7x = -14

<=> x = -2

Vậy x = -2