3/ Could you tell Johns about the school program by e-mail? (SEND)

-> Could you___________  John about the school program?

4/ Mai sent me a text message. (RECEIVED)

-> I_________Mai

5/ Did Huy say anything about your post on Facebook? (COMMENT)

-> Did Huy_________on Facebook?

1/ Remind me to send Minh a letter about our plans. (WRITE)

-> Remind me to ____________ our plans

2/ Some moneys use sign language to talk to people. (COMMUNICATION)

-> Some moneys____________ through sign language.

\(\sqrt{9-4\sqrt 5}+3\sqrt 5+11\\=\sqrt{4-4\sqrt 5+5}+3\sqrt 5+11\\=\sqrt{(2-\sqrt 5)^2}+3\sqrt 5+11\\=|2-\sqrt 5|+3\sqrt 5+11\\=\sqrt 5-2+3\sqrt 5+11\\=4\sqrt 5+9\)

Thay \(x=0,y=5\) vào hàm số (d) ta được:

\(5=0.m+5=5\) (luôn đúng)

\(\to\) (d) luôn đi qua A(0;5) với mọi m

ĐK: \(\begin{cases} x-2\sqrt{x-1}\ge 0\\x-1\ge 0\end{cases}\)

\(x-1\ge 0\\\leftrightarrow x\ge 1\\ x-2\sqrt{x-1}\ge 0\\\to x\ge 0\)

\(\to\) ĐKXĐ: \(x\ge 1\)

\(\sqrt{x-2\sqrt{x-1}}=2\\\leftrightarrow x-2\sqrt{x-1}=4\\\leftrightarrow x-2\sqrt{x-1}-4=0\\\leftrightarrow (x-1)-2\sqrt{x-1}+1-4=0\\\leftrightarrow (\sqrt{x-1}+1)^2-4=0\\\leftrightarrow (\sqrt{x-1}+1-2)(\sqrt{x-1}+1+2)=0\\\leftrightarrow (\sqrt{x-1}-1)(\sqrt{x-1}+3)=0\\\leftrightarrow \sqrt{x-1}-1=0\quad or\quad \sqrt{x-1}+3=0(\text{vô lý}\\\leftrightarrow \sqrt{x-1}=1\\\leftrightarrow x-1=1\\\leftrightarrow x=2(TM)\)

Vậy pt có nghiệm \(S=\{2\}\)

a/ Pt có 2 nghiệm phân biệt

\(\to\Delta'=(-m)^2-1.(-8m-16)=m^2+8m+16=(m+4)^2>0\\\to m+4>0\quad or\quad m+4<0\\\to m>-4\quad or\quad m<-4\)

b/ Theo Vi-ét:

\(\begin{cases}x_1+x_2=2m\\x_1x_2=-8m-16\end{cases}\)

\(x_1^2+x_2^2=5\\\leftrightarrow x_1^2+2x_1x_2+x_2^2-2x_1x_2=5\\\leftrightarrow (x_1+x_2)^2-2x_1x_2=5\\\leftrightarrow (2m)^2-2.(-8m-16)=5\\\leftrightarrow 4m^2+16m+32=5\\\leftrightarrow 4(m^2+4m+8)=5\\\leftrightarrow 4(m+2)^2+16=5\\\leftrightarrow 4(m+2)^2+11=0(\text{vô lý})\\\to m\in\varnothing\)

Vậy không có giá trị m thỏa mãn

a/ \(4-5x=0\\\leftrightarrow 5x=4\\\leftrightarrow x=\dfrac{4}{5}\)

Vậy nghiệm của đa thức f(x) là \(\dfrac{4}{5}\)

b/ Vì \(x^2\ge 0\\\to x^2+4\ge 0+4>0\\\to x^2+4>0\ne 0\)

\(\to\) Pt không có nghiệm

Vậy đa thức g(x) không có nghiệm

\(x>0\to \dfrac{4}{x}>0\)

Áp dụng BĐT Cô-si với 2 số dương \(x;\dfrac{4}{x}\)

\(\to x.\dfrac{4}{x}\ge 2\sqrt{x.\dfrac{4}{x}}=2.\sqrt 4=4\)

\(\to\) Dấu "=" xảy ra khi \(x=\dfrac{4}{x}\)

\(\to x=2\)

Ủa bạn là ai đấy

\(\bigg(\dfrac{1}{\sqrt x-1}-\dfrac{1}{\sqrt x+1}\bigg):\dfrac{x+1}{x-1}\\=\bigg(\dfrac{\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}-\dfrac{\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}\bigg.\dfrac{x-1}{x+1}\\=\dfrac{\sqrt x+1-\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{(\sqrt x-1)(\sqrt x+1)}{x+1}\\=\dfrac{2}{x+1}\)