`@` `\text {Ans}`

`\downarrow`

`(x-3)^2 - x^2 + 10x - 7`

`= x^2 - 6x + 9 - x^2 + 10x - 7`

`= (x^2 - x^2) + (-6x + 10x) + (9-7)`

`= 4x + 2`

\(\left(x-3\right)^2-x^2+10x-7\)

\(=\left(x^2-6x+9\right)-x^2+10x-7\)

\(=\left(x^2-x^2\right)-\left(6x-10x\right)+\left(9-7\right)\)

\(=4x+2\)

\(=2\left(2x+1\right)\)

Ta có: xOt + xOz = 180 ( 2 góc kề nhau )

Mà xOz = \(\dfrac{1}{2}\) xOt

=> xOt + \(\dfrac{1}{2}\) xOt = 180

=> \(\dfrac{3}{2}\) xOt = 180

=> xOt = 120

=> xOz = \(\dfrac{1}{2}\) xOt = \(\dfrac{1}{2}\) 120 = 60

=> zOy = 120 ( góc đối với xOt )

=> yOt = 120 ( góc đối với xOz )

 

So sánh:

\(-\dfrac{173}{457}\)  và  \(-\dfrac{16}{47}\)

\(\dfrac{-173}{457}=\dfrac{-173\times47}{457\times47}=\dfrac{-8131}{21479}\)

\(\dfrac{-16}{47}=\dfrac{-16\times457}{47\times457}=\dfrac{-7312}{21479}\)

Vì: \(-8131< -7312\)

=> \(\dfrac{-173}{457}< \dfrac{-16}{47}\)

a) Cho đa thức : \(x^2-5x+4=0\)

\(=>\left(x^2-x\right)-\left(4x-4\right)=0\\ =>x\left(x-1\right)-4\left(x-1\right)=0\\ =>\left(x-1\right)\left(x-4\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy nghiệm đa thức trên là : `x=1` hoặc `x=4`

b) Ta thấy : \(x^2+x+3=\left(x^2+\dfrac{1}{2}x\right)+\left(\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{11}{4}\\ =x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{11}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\in R\)

Vậy đa thức trên vô nghiệm

 Ta có \(-2x+3y\) \(=3\left(7x+y\right)-23x\), lại có \(7x+y⋮23\)  và \(23x⋮23\) nên \(3\left(7x+y\right)-23x⋮23\) hay \(-2x+3y⋮23\) (đpcm)

Theo đề ta có:

\(4a-8=3a+6\)

\(\Rightarrow4a-3a=6+8\)

\(\Rightarrow a=14\)

Vậy với a=14 thì f(a)=g(a)

Lời giải:
$x^2+5x+5=0$

$(x^2+2.2,5x+2,5^2)-1,25=0$

$(x+2,5)^2=1,25$

$\Rightarrow x+2,5=\pm \sqrt{1,25}$

$\Rightarrow x=\pm \sqrt{1,25}-2,5$

`@` `\text {Ans}`

`\downarrow`

\(\left(-3\right)^x\div81=-27\)

`=>`\(\left(-3\right)^x=-27\cdot81\)

`=>`\(\left(-3\right)^x=\left(-3\right)^3\cdot3^4\)

`=>`\(\left(-3\right)^x=\left(-3\right)^7\)

`=> x=7`

Vậy, `x=7.`

`@` `\text {Kaizuu lv u}`

(-3)^x . 81 = -27

<=> (-3)^x . (-3)⁴ = (-3)³

<=> (-3)^x+4 = (-3)³

=> x+4=3

<=>x= -1

a) Ta có: \(x\in B\left(6\right);15< x\le54\)

\(\Rightarrow x\in\left\{18;24;30;36;42;48;54\right\}\)

b) Ta có: \(x\in BC\left(5;8\right);40\le x\le200\)

\(\Rightarrow x\in\left\{40;80;120;160;200\right\}\)

c) Ta có: \(x\inƯC\left(36;48\right);4\le x\le20\)

\(\Rightarrow x\in\left\{4;6;9;12\right\}\)

d) Ta có: \(x\inƯC\left(18,36\right);2\le x\le30\)

\(\Rightarrow x\in\left\{2;3;6;9;18\right\}\)