a) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\left(đk:x\ne-5\right)\)

\(\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(\Rightarrow x=37\left(tm\right)\)

b) \(\dfrac{x^2}{6}=\dfrac{24}{49}\)

\(\Rightarrow x^2=\dfrac{24.6}{49}=\dfrac{144}{49}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{7}\\x=-\dfrac{12}{7}\end{matrix}\right.\)

D

a: \(2^{300}=8^{100}\)

\(3^{200}=9^{100}\)

mà 8<9

nên \(2^{300}< 3^{200}\)

Câu hỏi đâu bạn

bạn cần giúp gì???

\(A=\dfrac{2^{10}\cdot3^8}{2^{10}\cdot3^8\cdot2^8\cdot3^8\cdot2^2\cdot5}+\dfrac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{16}\cdot3^{12}+2^{18}\cdot3^{18}}\\ A=\dfrac{1}{2^{10}\cdot3^8\cdot5}+\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{16}\cdot3^{12}\left(1+2^2\cdot3^6\right)}\\ A=\dfrac{1}{2^{10}\cdot3^8\cdot5}+\dfrac{6}{2^4\cdot3^2\cdot2917}=\dfrac{1}{2^{10}\cdot3^8\cdot5}+\dfrac{1}{2^3\cdot3\cdot2917}\)

D

Bài 6:

a) \(x^2-2x+4=\left(x^2-2x+1\right)+3=\left(x-1\right)^2+3>0\forall x\)

b) \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1< 0\forall x\)

c) \(\left(x-2\right)\left(x-4\right)+3=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2>0\forall x\)

d) \(-2x^2+5x-19=\dfrac{-4x^2+10x-38}{2}=\dfrac{-\left(4x^2-10x+6,25\right)-31,75}{2}=\dfrac{-\left(2x-2,5\right)^2-31,75}{2}< 0\forall x\)

Câu 4:

a) \(x^5-x^3-x^2+1=\left(x^5-x^3\right)-\left(x^2-1\right)=x^3\left(x^2-1\right)-\left(x-1\right)\left(x+1\right)=x^3\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x+1\right)\left(x^3-1\right)=\left(x-1\right)^2\left(x+1\right)\left(x^2+x+1\right)\)

Câu 5:

\(a^3+b^3=3ab-1\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3ab+1=0\\ \Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b+1\right)=0\\ \Leftrightarrow\left(a+b+1\right)\left(a^2+b^2+1-ab-a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b+1=0\left(vô.lí.do.a,b>0\right)\\a^2+b^2+1-ab-a-b=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b=0\\ \Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)

Vậy \(T=\left(1-2\right)^{2020}+\left(1-1\right)^{2021}=\left(-1\right)^{2020}+0=1\)