1 . a) dấu hiệu ở đây khối lượng của 20 học sinh lớp 7 
 -giá trị ở đây là 8
b)

X= 30.2+32.3+33.1+34.2+35.1+ 36.6+ 38.4+46.1  : 20
  =  \(\dfrac{706}{20}\)_=35,3

a: =x^2+6x+9+x^2-6x+9+2x^2-32

=4x^2-14

b: =(x+3-10+x)^2=(2x-7)^2=4x^2-28x+49

c: =(x-3-x+5)^2=2^2=4

e: =x^2+10x+25-x^2+10x-25=20x

d: A=(5-1)(5+1)(5^2+1)(5^4+1)/4

=(5^2-1)(5^2+1)(5^4+1)/4

=(5^4-1)(5^4+1)/4

=(5^8-1)/4

g: =x^2-9-x^2-4x+5

=-4x-4

b)\(\sqrt{x^2-10x+25}=2x-3\)                                               ĐK:x≥3/2

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2x-3\)

\(\Leftrightarrow\left|x-5\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x-3\\x-5=3-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)

Vậy phương trình có 2 nghiệm là ... 

a: Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\)

\(\Leftrightarrow5\sqrt{x+3}-4\sqrt{x+3}=2\)

\(\Leftrightarrow x+3=4\)

hay x=1

c)ĐK:x≥1/2

\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-\sqrt{2x-1}+1}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)

\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)

\(\Leftrightarrow\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)  (do \(\sqrt{2x-1}+1\ge0\))

\(\Leftrightarrow\sqrt{2x-1}+1+\left|1-\sqrt{2x-1}\right|=2\)

Ta có:\(\sqrt{2x-1}+1+\left|1-\sqrt{2x-1}\right|\ge\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\)

                        \(\Leftrightarrow\dfrac{1}{2}\le x\le1\)

Vậy phương trình có tâp nghiệm \(S=\left\{x|\text{​​}\dfrac{1}{2}\le x\le1\right\}\)

26C 27A 28D 29A 30C 31D 32A 33D 34D 35A

a: \(1+tan^2a\)

\(=1+\dfrac{sin^2a}{cos^2a}=\dfrac{cos^2a+sin^2a}{cos^2a}=\dfrac{1}{cos^2a}\)

b: \(1+cot^2a=1+\dfrac{cos^2a}{sin^2a}\)

\(=\dfrac{sin^2a+cos^2a}{sin^2a}=\dfrac{1}{sin^2a}\)

c: \(cot^2a-cos^2a=\dfrac{cos^2a}{sin^2a}-cos^2a\)

\(=cos^2a\left(\dfrac{1}{sin^2a}-1\right)\)

\(=cos^2a\cdot\dfrac{1-sin^2a}{sin^2a}=\dfrac{cos^2a}{sin^2a}\cdot cos^2a=cot^2a\cdot cos^2a\)

d: \(\left(1+cosa\right)\left(1-cosa\right)=1-cos^2a=sin^2a\)

=>\(\dfrac{1+cosa}{sina}=\dfrac{sina}{1-cosa}\)

\(\left(a^2+ab+b^2\right)\cdot\left(a^2-ab+b^2\right)-\left(a^4+b^4\right)\)

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

\(=a^2b^2\)