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Bài 5:

a) \(\left(x-4\right)\left(x^2+4x+6\right)-x\left(x^2-6\right)=2\)

\(\Rightarrow x\left(x^2+4x+6\right)-4\left(x^2+4x+6\right)-x^3+6x=2\\ \Rightarrow x^3+4x^2+6x-4x^2-16x-24-x^3+6x=2\\ \Rightarrow-4x-24=2\\ \Rightarrow-4x=26\\ \Rightarrow x=-\dfrac{13}{2}\)

Bài 6:

\(a^3+b^3+3ab\\ =\left(a+b\right)\left(a^2+ab+b^2\right)+3ab\\ =a^2+ab+b^2+3ab\\ =a^2+4ab+b^2\)

GTNN (A)=3178+2017 khi x=0 ko co GTLN

GTLN(b)=2017 khi x=-3 va y=5 khong co GTNN

GTNN(c)=2018 khi x=-1 va y=5 khong co GTLN

neu can giai thich thi h

ko thi thoi 

em cũng muốn làm phước giúp chị lắm chứ nhưng em mới ở lớp 6 thui

giải thích ik kich ròi

sai đề

\(ĐK:x\ge5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{x-5}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow4b^2-3a^2=x-20\)

\(PT\Leftrightarrow4b^2-3a^2+a+b+ab=0\\ \Leftrightarrow4ab+4b^2-3a^2-3ab+a+b=0\\ \Leftrightarrow4b\left(a+b\right)-3a\left(a+b\right)+\left(a+b\right)=0\\ \Leftrightarrow\left(a+b\right)\left(4b-3a+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=0\left(\text{loại do }a+b>0\right)\\4b-3a+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow4\sqrt{x-5}=3\sqrt{x}-1\\ \Leftrightarrow16x-80=9x-6\sqrt{x}+1\\ \Leftrightarrow7x+6\sqrt{x}-81=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-\dfrac{27}{7}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=9\left(nhận\right)\)

a) Ta có: 3a+1<3b+1

\(\Leftrightarrow3a< 3b\)

hay a<b

b) BPT \(\Leftrightarrow-2a>-2b\) \(\Leftrightarrow a< b\)

mk lớp 6

trang6 27trang7 29 trang8 40 trang9 77 trang10 67

a)\(\dfrac{24}{36}\)=\(\dfrac{8}{12}\)

b)\(\dfrac{14}{56}\)=\(\dfrac{1}{4}\)

c)\(\dfrac{9}{24}\)=\(\dfrac{21}{56}\)

Chúc bạn học tốt!

a: x/12=24/36

nên x/12=2/3

=>x=2/3x12=8

b: 1/x=14/56

nên 1/x=1/14

hay x=14

c: x/56=9/24

=>x/56=3/8

hay x=21

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