\(ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow3\sqrt{2x-3}-2\sqrt{2x-3}+6\sqrt{2x-3}=1\\ \Leftrightarrow7\sqrt{2x-3}=1\\ \Leftrightarrow\sqrt{2x-3}=\dfrac{1}{7}\\ \Leftrightarrow2x-3=\dfrac{1}{49}\Leftrightarrow x=\dfrac{74}{49}\left(tm\right)\)

\(ĐK:x\ne\pm2\\ a,B=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\\ B=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}=x-1\\ b,B=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\\ c,B>0\Leftrightarrow x-1>0\Leftrightarrow x>1;x\ne2\)

\(=\left(x-3y\right)^2-25=\left(x-3y-5\right)\left(x-3y+5\right)\)

\(\left\{{}\begin{matrix}\widehat{ABP}=\widehat{MBC}\left(=\widehat{ABC}+90^0\right)\\BA=BM\\BP=BC\end{matrix}\right.\Rightarrow\Delta BAP=\Delta BMC\left(c.g.c\right)\\ \Rightarrow AP=CM;\widehat{BAP}=\widehat{BMC}\)

Gọi \(\left\{O\right\}=AP\cap CM\)

\(\widehat{AIO}=\widehat{BIM}\left(đđ\right)\\ \Rightarrow\widehat{AOI}=180^0-\left(\widehat{BAP}+\widehat{AIO}\right)=180^0-\left(\widehat{BMC}+\widehat{BIM}\right)=90^0\)

Lại có HD,DE,EG lần lượt là đtb \(\Delta ACD,\Delta ACM,\Delta APM\)

Do đó \(\left\{{}\begin{matrix}HD\text{//}AP;HD=\dfrac{1}{2}AP\left(1\right)\\DE\text{//}CM;DE=\dfrac{1}{2}CM\left(2\right)\\EG\text{//}AP;EG=\dfrac{1}{2}AP\left(3\right)\end{matrix}\right.\)

\(\left(1\right)\left(3\right)\Rightarrow HD\text{//}EG;HD=EG\\ \Rightarrow DEGH\text{ là hbh}\\ \text{Mà }AP=CM\Rightarrow HD=HE\\ \Rightarrow DEGH\text{ là hình thoi}\)

Mặt khác: \(DE\text{//}CM;AP\bot CM\Rightarrow AP\bot DE\)

Mà \(HD\text{//}AP\Rightarrow DE\text{//}HD\)

Vậy DEGH là hình vuông

\(\Rightarrow 2^x-26=6\\ \Rightarrow 2^x=32=2^5\\ \Rightarrow x=5\)

\(a,=y\left(x^2-2xy+y^2\right)=y\left(x-y\right)^2\\ b,=x^2\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(x^2-1\right)=\left(x-1\right)\left(x-2\right)\left(x+1\right)\)

Có 7 bạn thích ăn bánh mì

\(a,ĐK:x\ne\pm1\\ b,A=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\\ c,x=-2\Leftrightarrow A=\dfrac{-2+1}{-2-1}=\dfrac{-1}{-3}=\dfrac{1}{3}\)

Gọi số thứ nhất là a

Số thứ 2 là a-3

Số thứ 3 là a+2

Ta có \(a+a-3+a+2=125\)

\(\Rightarrow3\times a-1=125\\ \Rightarrow a\times3=126\\ \Rightarrow a=42\)

Vậy 3 số lần lượt là 42,39,45

Gọi số ảnh khối 6,7,8,9 lần lượt là \(a,b,c,d(a,b,c,d\in \mathbb{N^*})\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{8}=\dfrac{d}{10}=\dfrac{a+b+c+d}{2+5+8+10}=\dfrac{250}{25}=10\\ \Rightarrow\left\{{}\begin{matrix}a=20\\b=50\\c=80\\d=100\end{matrix}\right.\)