TOÁN LỚP 6 NHA MÌNH GHI NHẦM

1.-12(x-5)+7(3-x)=5.

#-12x+60+21-7x=5

#-12x-7x=5-60-21

#-19x=-76

#x=-76:(-19)

#x=4(TMĐK:x€Z)

Vậy x=4

#là dấu suy ra nhé! Máy mình không có dấu suy ra!

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3x-1\\t=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+2x-3}{-2}=1-x\\t=\dfrac{1-2x+3}{-2}=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

a: \(6^x=5\)

=>\(x=log_65\)

b: \(7^{3-x}=5\)

=>\(3-x=log_75\)

=>\(x=3-log_75\)

c: \(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\)

=>\(\left(\dfrac{3}{5}\right)^{x-2}=\left(\dfrac{3}{5}\right)^3\)

=>x-2=3

=>x=5

d: \(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\)

=>\(\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)

=>x=-1

a.

\(6^x=5\Rightarrow x=log_65\)

b.

\(7^{3-x}=5\Rightarrow3-x=log_75\)

\(\Rightarrow x=3-log_75\)

c.

\(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\Rightarrow x-2=log_{\dfrac{3}{5}}\left(\dfrac{27}{125}\right)\)

\(\Rightarrow x-2=3\Rightarrow x=5\)

d.

\(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\Rightarrow\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)

\(\Rightarrow x=-1\)

Bài 1:

a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}\)

b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)

c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6\)

d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}\)

\(=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10\)

Bài 2:

\(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)

\(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)

3.

\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)

\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)

\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)

Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)

4.

\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)

5.

\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)

\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)

1.

\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)

Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)

\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)

\(\Rightarrow m=\left\{-1;0;1;2\right\}\)

2.

\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)

Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow m=\left\{1;2;3;4\right\}\)

\(A=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow A_{min}=\frac{3}{4}\) khi \(x=\frac{1}{2}\), \(A_{max}\) ko tồn tại

\(B=x^2+2x+5=\left(x+1\right)^2+4\ge4\)

\(\Rightarrow B_{min}=4\) khi \(x=-1\) , \(B_{max}\) ko tồn tại

\(C=\left(5-x\right)^2+2017\ge2017\)

\(\Rightarrow C_{min}=2017\) khi \(x=5\) , \(C_{max}\) ko tồn tại

\(D=20-\left(2x+1\right)^2\le20\)

\(\Rightarrow D_{max}=20\) khi \(x=-\frac{1}{2}\), \(D_{min}\) ko tồn tại

\(E=x^2+2\left|y+1\right|+7\ge7\)

\(\Rightarrow E_{min}=7\) khi \(\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\) , \(E_{max}\) ko tồn tại

GIÚP MK NHÂ