3: 

a: \(\Leftrightarrow x+1-6\sqrt{x+1}-9=0\)

=>\(\left(\sqrt{x+1}-3\right)=0\)

=>x+1=9

=>x=8

b: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\left(\sqrt{\dfrac{1}{2}x+1}+3\right)}}=10\)

=>\(\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}-\dfrac{21}{4}}=10\)

=>\(\dfrac{1}{2}x-\dfrac{21}{4}-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}=100\)

=>\(\dfrac{7}{4}\cdot\sqrt{\dfrac{1}{2}x+1}=\dfrac{1}{2}x-\dfrac{21}{4}-100=\dfrac{1}{2}x-\dfrac{421}{4}\)

=>\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{2}{7}x-\dfrac{421}{7}\)

=>1/2x+1=(2/7x-421/7)^2

=>1/2x+1=4/49x^2-1684/49x+177241/49

=>\(x\simeq249,77;x\simeq177,36\)

Có BCNN(a,b).UCLN(a,b)= ab

=>60 . UCLN(a,b) = 180 

=> UCLN(a,b)=3

Giả sửd= UCLN(a,b) ( d khác 0 )

có a=dm, b = dn 

ab= 180 => dmdn=180 => mn = 180 : (3.3) => mn=20=1.20=2.10=4.5

Ta có bảng sau 

vậy : (a,b)=(3;60),(6;30),(15;12),(12;15),(30;6),(6;30)

Bạn cần giúp bài nào nhỉ?

Bài giải

Vì : 1 kg = 100 dag

nên 4 kg = 400 dag

400 + 8 = 408 dag

Đáp án : 4 kg 8 dag = 408 dag

408 dag nha!

4kg 8dag=408dag

Hok tốt

1.

Phương trình có 2 nghiệm dương pb khi:

\(\left\{{}\begin{matrix}\Delta'=\left(m+1\right)^2-\left(2m+46\right)=m^2-45>0\\x_1+x_2=2\left(m+1\right)>0\\x_1x_2=2m+46>0\end{matrix}\right.\) \(\Rightarrow m>3\sqrt{5}\)

Khi đó:

\(\left|\sqrt{x_1}-\sqrt{x_2}\right|=2\)

\(\Leftrightarrow x_1+x_2-2\sqrt{x_1x_2}=4\)

\(\Leftrightarrow2\left(m+1\right)-2\sqrt{2m+46}=4\)

\(\Leftrightarrow2m+46-2\sqrt{2m+46}-48=0\)

Đặt \(\sqrt{2m+46}=a>0\)

\(\Rightarrow a^2-2a-48=0\Leftrightarrow\left[{}\begin{matrix}a=8\\a=-6\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2m+46}=8\)

\(\Rightarrow m=9\)

2.

Kết hợp pt thứ 2 và điều kiện đề bài ta được:

\(\left\{{}\begin{matrix}mx+3y=m+3\\x-3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(m+1\right)x=m+5\\x-3y=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m\ne-1\\x=\dfrac{m+5}{m+1}\\y=\dfrac{-m+3}{3\left(m+1\right)}\end{matrix}\right.\)

Thay vào pt đầu:

\(\Rightarrow\dfrac{2\left(m+5\right)}{m+1}+\dfrac{\left(m-1\right)\left(-m+3\right)}{3\left(m+1\right)}=4\)

\(\Rightarrow m^2-2m-15=0\Rightarrow\left[{}\begin{matrix}m=-5\\m=3\end{matrix}\right.\)

ảnh lỗi

a: \(VP=a^3+b^3+c^3-3bac\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VT\)

b: \(VT=\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)\)

\(=3a^2+15a+2ab+10b-a-5-2ab+4b\)

\(=3a^2+14a+14b-5\)

\(VP=\left(3a+5\right)\left(a+3\right)+2\left(7b-10\right)\)

\(=3a^2+9a+5a+15+14b-20\)

\(=3a^2+14a+14b-5\)

=>VT=VP

c: \(VT=a\left(b-x\right)+x\left(a+b\right)\)

\(=ab-ax+ax+bx\)

\(=ab+bx=b\left(a+x\right)=VP\)

d: \(VT=a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)

\(=ab-ac-ab-bc+ca-cb\)

\(=-2bc\)

=VP