\(5x\left(x-2\right)-\left(2-x\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{5}\end{matrix}\right.\)

`5x(x-2)-(2-x)=0`

`<=>5x(x-2)+(x-2)=0`

`<=>(x-2)(5x+1)=0`

`=>x-2=0` hoặc `5x+1=0`

`<=>x=2` hoặc `x=-1/5`

`12x^2 - 4x =4x(3x-1)`.

`->` Đúng

sai

Ta có :

 \(VT=x\left(y-1\right)+3\left(y-1\right)=\left(y-1\right)\left(x+3\right)\)

\(VP=-\left(1-y\right)\left(x+3\right)=\left(y-1\right)\left(x+3\right)\)

\(\Rightarrow VT=VP\)

Vậy đẳng thức trên đúng

(x+3) y-x+3=(4x+9)y-4x-9

-(4x+9) y+(x+3) y-x-(-4x)+12=0

-3((x-2) y-x-4)=0

(x+2) y-x-4=0

x+2=0

x=-2

y-1=0

y=1

\(14x^2y-21xy^2+28x^2y^2\\ =7xy\left(2x-3y+4xy\right)\)

`14x^2y - 21xy^2 + 28x^2y^2`

`= 7xy(2x-3y+4xy)`.

`3x-12x^{2}`

`=3x.1-3x.4x`

`=3x.(1-4x)`

Lời giải:

$3x-12x^2=3x(1-4x)$

Đề thiếu. Bạn xem lại.

   (2x³ - 3x - 1).(5x +2)

= 10x⁴ +  4x³ - 15x² - 6x - 5x - 2

= 10x⁴ + 4x³ - 15x² - 11x - 2

   (2x3 - 3x - 1).(5x +2)

= 10x4 +  4x3 - 15x2 - 6x - 5x - 2

= 10x4 + 4x3 - 15x2 - 11x - 2

Đáp án:

$\widehat{BKC}={110}^o$

Giải thích các bước giải:

a) Ta có:

$K$ đối xứng với $H$ qua $BC$

$\Rightarrow BC$ là trung trực của $HK$

$\Rightarrow BH=BK;CH=CK$

Xét $\Delta BHC$ và $\Delta BKC$ có:

$BH=BK\left(cmt\right)$

$CH=CK\left(cmt\right)$

$BC:$ cạnh chung

Do đó $\Delta BHC=\Delta BKC(c.c.c)$

b) Ta có:

$\widehat{BHK}=\widehat{BAH}+\widehat{ABH}$ (góc ngoài của $\Delta ABH$)

$\widehat{CHK}=\widehat{CAH}+\widehat{ACH}$ (góc ngoài của $\Delta ACH$)

$\Rightarrow \widehat{BHC}=\widehat{BHK}+\widehat{CHK}$

$=\widehat{BAH}+\widehat{ABH}+\widehat{CAH}+\widehat{ACH}$

$=\widehat{BAC}+\widehat{ABH}+\widehat{ACH}$

Ta lại có:

$\widehat{BAC}+\widehat{ABH}={90}^o$ $(BH\perp AC)$

$\widehat{BAC}+\widehat{ACH}={90}^o$ $(CH\perp AB)$

$\Rightarrow 2\widehat{BAC}+\widehat{ABH}+\widehat{ACH}={180}^o$

$\Rightarrow \widehat{ABH}+\widehat{ACH}={180}^o-2\widehat{BAC}$

Do đó:

$\widehat{BHC}=\widehat{BAC}+{180}^o-2\widehat{BAC}={180}^o-\widehat{BAC}={180}^o-{70}^o={110}^o$

Mặt khác:

$\widehat{BHC}=\widehat{BKC}(\Delta BHC=\Delta BKC)$

$\Rightarrow \widehat{BKC}={110}^o$

Đáp án:

$\widehat{BKC}={110}^o$

Giải thích các bước giải:

a) Ta có:

$K$ đối xứng với $H$ qua $BC$

$\Rightarrow BC$ là trung trực của $HK$

$\Rightarrow BH=BK;CH=CK$

Xét $\Delta BHC$ và $\Delta BKC$ có:

$BH=BK\left(cmt\right)$

$CH=CK\left(cmt\right)$

$BC:$ cạnh chung

Do đó $\Delta BHC=\Delta BKC(c.c.c)$

b) Ta có:

$\widehat{BHK}=\widehat{BAH}+\widehat{ABH}$ (góc ngoài của $\Delta ABH$)

$\widehat{CHK}=\widehat{CAH}+\widehat{ACH}$ (góc ngoài của $\Delta ACH$)

$\Rightarrow \widehat{BHC}=\widehat{BHK}+\widehat{CHK}$

$=\widehat{BAH}+\widehat{ABH}+\widehat{CAH}+\widehat{ACH}$

$=\widehat{BAC}+\widehat{ABH}+\widehat{ACH}$

Ta lại có:

$\widehat{BAC}+\widehat{ABH}={90}^o$ $(BH\perp AC)$

$\widehat{BAC}+\widehat{ACH}={90}^o$ $(CH\perp AB)$

$\Rightarrow 2\widehat{BAC}+\widehat{ABH}+\widehat{ACH}={180}^o$

$\Rightarrow \widehat{ABH}+\widehat{ACH}={180}^o-2\widehat{BAC}$

Do đó:

$\widehat{BHC}=\widehat{BAC}+{180}^o-2\widehat{BAC}={180}^o-\widehat{BAC}={180}^o-{70}^o={110}^o$

Mặt khác:

$\widehat{BHC}=\widehat{BKC}(\Delta BHC=\Delta BKC)$

$\Rightarrow \widehat{BKC}={110}^o$