Phương trình bậc nhất một ẩn

Giải thích

*BM cắt DC tại F.

△ABM có: AB//DF \(\Rightarrow\dfrac{BM}{MF}=\dfrac{AM}{MD}=\dfrac{AM}{2AM}=\dfrac{1}{2}\) (định lí Ta-let)

△BCF có: \(\dfrac{BM}{MF}=\dfrac{BN}{NC}=\dfrac{1}{2}\Rightarrow\)MN//BC//AB (định lí Ta-let đảo).

△ADC có: ME//DC

 \(\Rightarrow\dfrac{ME}{DC}=\dfrac{AM}{AD}=\dfrac{AM}{AM+MD}=\dfrac{AM}{AM+2AM}=\dfrac{1}{3}\) (hq định lí Ta-let)

\(\Rightarrow ME=\dfrac{DC}{3}=\dfrac{3}{3}=1\left(cm\right)\)

Chọn B

\(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}=x+3+\left|x-3\right|=x+3+3-x=6\)

\(x+3+\sqrt{x^2-6x+9}\)

\(=x+3+\sqrt{\left(x-3\right)^2}\)

\(=x+3+\left|x-3\right|=x+3+\left|3-x\right|\)

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

|x+3| = 2x+3 (đk : \(-\dfrac{3}{2}\)) 
xét 2 th 
th1 :x+3 = 2x +3 
<=> x - 2x = 3 - 3 
<=> -x = 0 
<=> x = 0 
th2 : x + 3 = -2x-3
<=> x+2x = -3 -3 
<=> 3x = -6 
<=> x = -2

\(\left(2x+m\right)\left(x-1\right)-2x^2+mx-2=0\)

\(\Leftrightarrow2x^2-2x+mx-m-2x^2+mx-2=0\)

\(\Leftrightarrow-2x+2mx-m-2=0\)

\(\Leftrightarrow2x\left(m-1\right)=m+2\)

\(\Leftrightarrow x=\dfrac{m+2}{2\left(m-1\right)}\)

Để phương trình có nghiệm là 1 số không âm thì:

\(\left\{{}\begin{matrix}m\ne1\\\dfrac{m+2}{2\left(m-1\right)}\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m+2\ge0\\2\left(m-1\right)\ge0\end{matrix}\right.hay\left\{{}\begin{matrix}m+2\le0\\2\left(m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ge-2\\m>1\end{matrix}\right.hay\left\{{}\begin{matrix}m\le-2\\m< 1\end{matrix}\right.\)

\(\Leftrightarrow m>1\) hay \(m\le-2\).

-Vậy \(m>1\) hay \(m\le-2\) thì phương trình có nghiệm là 1 số không âm.

*vn:vô nghiệm.

a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

Xét `\triangle ABC` có: `\hat{A}+\hat{B}+\hat{C}=180^o`

         Mà `\hat{A}=2\hat{B}=>\hat{B}=1/2\hat{A}`

               `\hat{C}=\hat{A}+30^o`

 `=>\hat{A}+1/2\hat{A}+\hat{A}+30^o = 180^o`

 `=>5/2\hat{A}=150^o`

 `=>\hat{A}=60^o`

Gọi x là số đo góc A.

       y là số đo góc B.

       z là số đo góc C.

Theo đề: \(x=2y\Leftrightarrow y=\dfrac{x}{2}\)

\(c=30+x\)

Mà: x + y + z = 180^o

\(\Rightarrow x+\dfrac{x}{2}+30+x=180^o\)

\(\Leftrightarrow2x+\dfrac{x}{2}=150^o\)

\(\Leftrightarrow\dfrac{4x}{2}+\dfrac{x}{2}=\dfrac{300^o}{2}\)

\(\Leftrightarrow5x=300^o\)

\(\Leftrightarrow x=60^o\)

Vậy số đo góc A là 60^o

\(=\dfrac{\left(x-1\right)\left(x-2\right)-x\left(x+2\right)+8}{x^2-4}\left(đk:x\ne\pm2\right)\)

\(=\dfrac{x^2-2x-x+2-x^2-2x+8}{x^2-4}\)

\(=\dfrac{-5x+10}{x^2-4}=\dfrac{-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{5}{x+2}\)

xem lại đề bài 

\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}+\dfrac{8}{x^2-4}=0\left(x\ne\pm2\right)\) (do ko có kết quả nên mik nghĩ là 0)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{x^2-4}-\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{8}{x^4-4}=0\)

\(\Rightarrow x^2-2x-x+2-x^2-2x+8=0\)

\(\Leftrightarrow-5x=-10\)

\(\Leftrightarrow x=2\left(loại\right)\)

Vậy pt có \(S=\varnothing\)

=>x^2-2x+6x+12=2x+12

=>x^2+4x-2x=0

=>x(x+2)=0

=>x=0(nhận) hoặc x=-2(loại)