dễ mà bạn

a)3x-18=0                     à mà mik chx hc phương trình 

3x=18+0                          sorry bạn nhé

3x=18

x=18:3

x=6

vậy x=6

a)\(3x-18=0\)

\(\Leftrightarrow3x=18\)

\(\Leftrightarrow x=6\)

Vậy x=6

b)\(2x.\left(x-4\right)-3x+12=0\)

\(\Leftrightarrow2x.\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(2x-3\right).\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=4\end{cases}}}\)

Vậy .......

c)\(\frac{x-1}{2}-\frac{x+3}{3}=x+1\)

\(\Leftrightarrow6.\left(\frac{x-1}{2}-\frac{x+3}{3}\right)=6.\left(x+1\right)\)

\(\Leftrightarrow3.\left(x-1\right)-2.\left(x+3\right)=6x+6\)

\(\Leftrightarrow3x-3-2x-6=6x+6\)

\(\Leftrightarrow3x-2x-6x=6+3+6\)

\(\Leftrightarrow-5x=15\)

\(\Leftrightarrow x=-3\)

Vậy x= -3

d)\(\frac{x-3}{x+3}-\frac{5}{3-x}=\frac{30}{x^2-9}\)

\(\Leftrightarrow\frac{x-3}{x+3}-\frac{-5}{x-3}=\frac{30}{\left(x+3\right).\left(x-3\right)}\)

\(\Leftrightarrow\frac{\left(x-3\right).\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}-\frac{-5.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}=\frac{30}{\left(x-3\right).\left(x+3\right)}\)

\(\Leftrightarrow\left(x-3\right)^2-\left(-5\right).\left(x+3\right)=30\)

\(\Leftrightarrow x^2-6x+9-\left(-5x-15\right)=30\)

\(\Leftrightarrow x^2-6x+9+5x+15-30=0\)

\(\Leftrightarrow x^2-x-6=0\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}}\)

Vậy......

a) Xét tam giác \(AEF\)và tam giác \(MEC\): 

\(\widehat{AEF}=\widehat{MEC}\)(đối đỉnh) 

\(\widehat{EAF}=\widehat{EMC}\left(=90^o\right)\)

suy ra \(\Delta AEF~\Delta MEC\left(g.g\right)\).

b)  Xét tam giác \(BAC\)và tam giác \(BMF\): 

\(\widehat{B}\)chung

\(\widehat{BAC}=\widehat{BMF}\left(=90^o\right)\)

suy ra \(\Delta BAC~\Delta BMF\left(g.g\right)\).

suy ra \(\frac{BA}{BM}=\frac{BC}{BF}\)

\(\Rightarrow BA.BF=BM.BC\)

c) \(BA.BF=BM.BC\Rightarrow\frac{BA}{BC}=\frac{BM}{BF}\)

Xét tam giác \(BAM\)và tam giác \(BCF\): 

\(\frac{BA}{BC}=\frac{BM}{BF}\)

\(\widehat{B}\)chung

suy ra \(\Delta BAM~\Delta BFC\left(c.g.c\right)\)

\(\Rightarrow\widehat{BMA}=\widehat{BFC}\).

d) Đặt \(\frac{BF}{BM}=\frac{BC}{BA}=k\)

suy ra \(\frac{S_{BAC}}{S_{ABM}}=k^2\)

mà \(S_{AMCF}=3S_{ABM}\Rightarrow S_{ABC}=4S_{ABM}\Rightarrow k=2\).

\(\frac{BC}{BA}=2\Rightarrow tanB=2\Rightarrow\widehat{B}=arctan\left(2\right).\)

cái chỗ trên cạnh AB và AC là AB và BC đk

minh cung chua hoc bai nay

Các bạn ơi, BF cắt AE tại G nhé.

`Answer:`

\(\frac{\left(m^2+1\right).x+1-2m^2}{x-5}\)\(=2m\left(ĐKXĐ:x\ne5\right)\)

`=>(m^2+1).x+1-2m^2=2m.(x-5)`

`<=>(m^2-2m+1).x=2m^2-10m-1`

`<=>(m-1)^2``x=2m^2-10m-1(1)`

Phương trình `(1)` có nghiệm duy nhất `<=>m\ne1`

Lúc đó nghiệm của phương trình `(1)` sẽ là \(x=\frac{2m^2-10m-1}{\left(m-1\right)^2}\)

Ta xét hiệu:

 \(x-3=\frac{2m^2-10m-1}{\left(m-1\right)^2}-3\)

\(=\frac{2m^2-10m-1-3\left(m-1\right)^2}{\left(m-1\right)^2}\)

\(=\frac{-m^2-4m-4}{\left(m-1\right)^2}\)

\(=-\frac{\left(m+2\right)^2}{\left(m-1\right)^2}\)

\(\le0;\forall m\ne1\)

\(\Rightarrow x-3\le0;\forall m\ne1\)

\(\Rightarrow x\le3;\forall m\ne1\)

Dấu "=" xảy ra khi `m+2=0<=>m=-2`

tt

liuliu

\(\dfrac{x-2}{40}-\dfrac{2-x}{102}+1=1-\dfrac{2x-4}{2022}\\ \Leftrightarrow\dfrac{x-2}{40}+\dfrac{x-2}{102}+\dfrac{x-2}{1011}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{1}{40}+\dfrac{1}{102}+\dfrac{1}{1011}\right)=0\\ \Leftrightarrow x-2=0\left(vì.\dfrac{1}{40}+\dfrac{1}{102}+\dfrac{1}{1011}\ne0\right)\\ \Leftrightarrow x=2\)

\(\Leftrightarrow4x^2-4xy+4y^2=12\)

\(\Leftrightarrow\left(2x-y\right)^2+3y^2=12\)

\(\Leftrightarrow3y^2=12-\left(2x-y\right)^2\le12\)

\(\Rightarrow y^2\le4\Rightarrow y^2=\left\{0;1;4\right\}\)

\(\Rightarrow y=\left\{-2;-1;0;1;2\right\}\)

Thế vào pt ban đầu tính x tương ứng

a, \(\Rightarrow x-5\ne0\Leftrightarrow x\ne5\)

b, mà \(\left(5x-3\right)^2\ge0\forall x\)

 \(\Rightarrow5x-3=0\Leftrightarrow x=\dfrac{3}{5}\)