Số quả đủ chia 1 người 4 quả

Số đào đủ chia cho 1 người 4 quả nhiều hơn số đào đủ chia 1 người 3 quả là:

2 + 2 = 4 (quả)

Một em được chia 4 quả nhiều hơn 1 em được chia 3 quả là: 

4 – 3 = 1 (quả)

Số em được chia đào là:

4 : 1 = 4 (em)

Số đào có là:

4 x 3 + 2 = 14 (quả)

Ta có: a+b+c=0

$\iff {\left(a+b+c\right)}^2=0$

$\iff a^2+b^2+c^2+2\left(ab+bc+ac\right)=0$

$\iff 2\left(ab+bc+ac\right)=0-1=-1$

hay $ab+bc+ac=-\genfrac{}{}{0ex}{}{1}{2}$

$\iff {\left(ab+bc+ac\right)}^2=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2+2a{b}^{2}c+2ab{c}^2+2a^{2}bc=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2+2abc\left(b+c+a\right)=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2=\genfrac{}{}{0ex}{}{1}{4}$

Ta có: $M=a^4+b^4+c^4$

$\iff M=a^4+b^4+c^4+2a^2{b}^2+2a^2{c}^2+2b^2{c}^2-2a^2{b}^2-2a^2{c}^2-2b^2{c}^2$

$\iff M={\left(a^2+b^2+c^2\right)}^2-2\left(a^2{b}^2+a^2{c}^2+b^2{c}^2\right)$

$\iff M=1^2-2\cdot \genfrac{}{}{0ex}{}{1}{4}=1-\genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{2}$

Vậy: $M=\genfrac{}{}{0ex}{}{1}{2}$

Ta có: a+b+c=0

$\iff {\left(a+b+c\right)}^2=0$

$\iff a^2+b^2+c^2+2\left(ab+bc+ac\right)=0$

$\iff 2\left(ab+bc+ac\right)=0-1=-1$

hay $ab+bc+ac=-\genfrac{}{}{0ex}{}{1}{2}$

$\iff {\left(ab+bc+ac\right)}^2=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2+2a{b}^{2}c+2ab{c}^2+2a^{2}bc=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2+2abc\left(b+c+a\right)=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2=\genfrac{}{}{0ex}{}{1}{4}$

Ta có: $M=a^4+b^4+c^4$

$\iff M=a^4+b^4+c^4+2a^2{b}^2+2a^2{c}^2+2b^2{c}^2-2a^2{b}^2-2a^2{c}^2-2b^2{c}^2$

$\iff M={\left(a^2+b^2+c^2\right)}^2-2\left(a^2{b}^2+a^2{c}^2+b^2{c}^2\right)$

$\iff M=1^2-2\cdot \genfrac{}{}{0ex}{}{1}{4}=1-\genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{2}$

Vậy: $M=\genfrac{}{}{0ex}{}{1}{2}$

Ta có: a+b+c=0

$\iff {\left(a+b+c\right)}^2=0$

$\iff a^2+b^2+c^2+2\left(ab+bc+ac\right)=0$

$\iff 2\left(ab+bc+ac\right)=0-1=-1$

hay $ab+bc+ac=-\genfrac{}{}{0ex}{}{1}{2}$

$\iff {\left(ab+bc+ac\right)}^2=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2+2a{b}^{2}c+2ab{c}^2+2a^{2}bc=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2+2abc\left(b+c+a\right)=\genfrac{}{}{0ex}{}{1}{4}$

$\iff a^2{b}^2+b^2{c}^2+a^2{c}^2=\genfrac{}{}{0ex}{}{1}{4}$

Ta có: $M=a^4+b^4+c^4$

$\iff M=a^4+b^4+c^4+2a^2{b}^2+2a^2{c}^2+2b^2{c}^2-2a^2{b}^2-2a^2{c}^2-2b^2{c}^2$

$\iff M={\left(a^2+b^2+c^2\right)}^2-2\left(a^2{b}^2+a^2{c}^2+b^2{c}^2\right)$

$\iff M=1^2-2\cdot \genfrac{}{}{0ex}{}{1}{4}=1-\genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{2}$

Vậy: $M=\genfrac{}{}{0ex}{}{1}{2}$