=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

Tách 1+2-3-4+5-6-8+...+97-98-99-100    với 101

A= 1 + 2 - 3 -4 + 5 + 6 -7 -8 + ... +97 +98 -99 -100 ( có: ( 100 - 1 ) : 1 + 1 = 100 )

                   A= ( 1 +2 - 3 - 4 ) + ( 5 + 6 - 7 -8 ) + ... ( 97 + 98 - 99 +100 ) ( có 100 : 4 = 25 cặp )

                   A= - 4 + -4 + -4 + ... + -4 ( có 25 số hạng )

                   A= ( -4 ) . 25

                   A= -100 + 101

A=1 

học tốt

TKS NGUYÊN NGOC LINH

1b) Ta có: \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{101}{100}=\frac{3.4.5....101}{2.3.4....100}=\frac{101}{2}\)

AI K CHO MU TA THẾ

MK NÈ, BẠN ẤY LÀM TỐT MÀ!

\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=2\sqrt{6}\cdot3\sqrt{6}-4\sqrt{3}\cdot3\sqrt{6}+5\sqrt{2}\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+30\sqrt{3}\)

1-2+3-4+...+2021-2022+2023

=(1-2)+(3-4)+...+(2021-2022)+2023

=(-1)+(-1)+(-1)+...+(-1)+2023

=(-1011)+2023

=1012

\(1-2+3-4+...+2021-2022+2023\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2023\)

\(=-1011+2023=1012\)

\(M=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+1\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+1\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)

\(=6+\sqrt{6}-11\sqrt{6}-11=-5-10\sqrt{6}\)

\(M=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)\)

\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+1\right)\)

\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+1\right)\)

\(M=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+1\right)\)

\(M=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\cdot\left(\sqrt{6}+1\right)\)

\(M=\left(5\sqrt{6}-4\sqrt{6}+1-12\right)\left(\sqrt{6}+1\right)\)

\(M=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)

\(M=6+\sqrt{6}-11\sqrt{6}-11\)

\(M=-10\sqrt{6}-5\)

x : y : z = 3 : 4 : 5

=>\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)

ADTCDTSBN:

\(\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}=\dfrac{2x^2+2y^2-3z^2}{18+32+75}=\dfrac{-4}{5}\)

\(\dfrac{x}{3}=\dfrac{-4}{5}\Rightarrow x=\dfrac{-12}{5}\)

\(\dfrac{y}{4}=\dfrac{-4}{5}\Rightarrow y=\dfrac{-16}{5}\)

\(\dfrac{z}{5}=\dfrac{-4}{5}\Rightarrow z=-4\)

\(x:y:z=3:4:5=>\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

\(=>x=\dfrac{3y}{4},z=\dfrac{5y}{4}\) thay x,z vào \(2x^2+2y^2-3z^2=-100\)

\(< =>2\left(\dfrac{3y}{4}\right)^2+2y^2-3\left(\dfrac{5y}{4}\right)^2=-100\)

\(=>y=\pm8\)

* với y=8 \(=>x=\dfrac{3.8}{4}=6,z=\dfrac{5.8}{4}=10\)

* với y=-8 \(=>x=-6,z=-10\)