Bài 1:

\(a^2+b^2+c^2=16\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=16\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=16\Rightarrow ab+bc+ac=-8\)\(\Rightarrow\left(ab+bc+ac\right)^2=64\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=64\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=64\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=64\)

Ta có:

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=16^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=256-2.64=128\)

Fan sơn tùng là đây

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow x^2=36\Rightarrow x=\pm6\)

\(\Rightarrow x^2=64\Rightarrow x=\pm8\)

Vậy .....

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

Áp dụng t.c của dãy tỉ só bằng nhau,ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{16}{7}\)

=>\(x=\dfrac{16}{7}.3=\dfrac{48}{7}\)

\(y=\dfrac{16}{7}.4=\dfrac{64}{7}\)

\(z=\dfrac{16}{7}.5=\dfrac{80}{7}\)

Vậy...

Các câu sau tương tự

ADTC dãy tỉ số bằng nhau ta cs

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x^2+y^2}{2^2+3^2}=\frac{52}{13}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=4\\\frac{y}{3}=4\\\frac{z}{4}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=16\end{matrix}\right.\)

Ta có:

\(2A=2+2^2+2^3+...+2^{101}\)

=>\(2A-A=\left(2+2^2+..+2^{101}\right)-\left(1+2+2^2+..+2^{100}\right)\)

=>\(A=2^{101}-1\)

Vì \(2^{101}-1>2^{100}-1\) nên A>B

Vậy A>B

Vì A có 2¹⁰⁰ và được cộng thêm, B có 2¹⁰⁰ phải trừ 1 nên A > B.

ngắn gọn thôi

a ) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)

Do \(2016^2>2016^2-1\)

\(\Rightarrow B>A\)

b ) \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1< 2^{32}=D\)

Vậy \(C< D\)

so sánh :a)A=2015.2017 va B=2016²

Ta có: A = 2015.2017 = (2016-1)(2016+1)

= 2016²-1<2016²

=> A < B

a) Đặt: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\) 

Ta có: \(x^2+3y^2-2z^2=-16\)

\(\Rightarrow\left(2k\right)^2+3\cdot\left(3k\right)^2-2\cdot\left(4k\right)^2=-16\)

\(\Rightarrow4k^2+3\cdot9k^2-2\cdot16k^2=-16\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow-k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

Với k = 4

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{3}=4\\\dfrac{z}{4}=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot4=8\\y=3\cdot4=12\\z=4\cdot4=16\end{matrix}\right.\)

Với k = -4 

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-4\\\dfrac{y}{3}=-4\\\dfrac{z}{4}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot-4=-8\\y=3\cdot-4=-12\\z=4\cdot-4=-16\end{matrix}\right.\) 

Vậy: ...

b) Đặt: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) 

Ta có: \(2x^2+2y^2-3z^2=-100\)

\(\Rightarrow2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)

\(\Rightarrow2\cdot9k^2+2\cdot16k^2-3\cdot25k^2=-100\)

\(\Rightarrow18k^2+32k^2-75k^2=-100\)

\(\Rightarrow-25k^2=-100\)

\(\Rightarrow k^2=-\dfrac{100}{-25}=4\)

\(\Rightarrow k=\pm2\)

Với k = 2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=2\\\dfrac{y}{4}=2\\\dfrac{z}{5}=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=2\cdot4=8\\z=2\cdot5=10\end{matrix}\right.\)

Với k = -2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-2\\\dfrac{y}{4}=-2\\\dfrac{z}{5}=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot-3=-6\\y=2\cdot-4=-8\\z=2\cdot-5=-10\end{matrix}\right.\)

Vậy: ...