\(\left|4-x\right|+2x=0\)

`TH1:x<=4`

`(4-x)+2x=0`

`=>4-x+2x=0`

`=>x+4=0`

`=>x=-4(tm)` 

`TH2:x>4`

`-(4-x)+2x=0`

`=>-4+x+2x=0`

`=>3x-4=0`

`=>3x=4`

`=>x=4/3` (ktm)

Vậy: ... 

\(x^2+2\left|x-\dfrac{1}{2}\right|=\left|x^2+2\right|\\ =>x^2+2\left|x-\dfrac{1}{2}\right|=x^2+2\left(x^2+2>0\right)\\ =>2\left|x-\dfrac{1}{2}\right|=2\\ =>\left|x-\dfrac{1}{2}\right|=1\\ TH1:x>\dfrac{1}{2}\\ =>x-\dfrac{1}{2}=1\\ =>x=1+\dfrac{1}{2}=\dfrac{3}{2}\\ TH2:x< =\dfrac{1}{2}\\ =>x-\dfrac{1}{2}=-1\\ =>x=\dfrac{1}{2}-1=-\dfrac{1}{2}\left(tm\right)\)

no

a.

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)

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

b.

Từ câu a:

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

\(\Rightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow a^4+b^4+c^4=\dfrac{\left(a^2+b^2+c^2\right)^2}{2}\)

                Giải:

Tỉ số số bi của An và số bi của Bình là:

          \(\dfrac{1}{7}\) : \(\dfrac{1}{2}\) = \(\dfrac{2}{7}\) 

Theo bài ra ta có sơ đồ:

Theo sơ đồ ta có:

Số bi của An là:

150 : (7 - 2) x 2 = 60 (viên bi)

Số bi của Bình là:

150 +  60 =  210 (viên bi)

Đáp số: .... 

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{5}\\ ===============\\ \left(1-\dfrac{3}{4}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times...\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\\ =\dfrac{1}{4}\times\dfrac{4}{7}\times\dfrac{7}{10}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\\ =\dfrac{1}{100}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)

\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)

\(=\dfrac{1}{5}\)

\(\left(1-\dfrac{3}{4}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{13}\right)\times...\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\)

\(=\dfrac{1}{4}\times\dfrac{4}{7}\times\dfrac{7}{10}\times\dfrac{10}{13}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\)

\(=\dfrac{1\times4\times7\times10\times...\times94\times97}{4\times7\times10\times13\times...\times97\times100}\)

\(=\dfrac{1}{100}\)

xy-x=3y+10

=>x(y-1)-3y=10

=>x(y-1)-3y+3=13

=>(x-3)(y-1)=13

=>\(\left(x-3;y-1\right)\in\left\{\left(1;13\right);\left(13;1\right);\left(-1;-13\right);\left(-13;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(4;14\right);\left(16;2\right);\left(2;-12\right);\left(-10;0\right)\right\}\)

Diện tích bức tường là 4x2=8(m²)

Thời gian chú An sơn xong bức tường là:

8x2=16(phút)

 A = (14⁸)²⁰²⁰ + 10 

A =  (14⁸)^5.404 + 10

A = (14⁵)^8.404 + 10

A = 537824³²³² + 10

537824 \(\equiv\) 1 (mod 11)

537824³²³² \(\equiv\) 1³²³² (mod 11) \(\equiv\) 1 (mod 11)

10 \(\equiv\) 10 (mod 11)

⇒ 537824³²³² + 10  \(\equiv\) 1 + 10 (mod 11)

⇒537824³²³² + 10 \(\equiv\) 11 (mod 11) \(\equiv\) 0 (mod 11)

⇒ A = (14⁸)²⁰²⁰ + 10 \(⋮\) 11 (đpcm)

\(14\equiv3\left(mod11\right)\Rightarrow\left(14^8\right)^{2020}\equiv\left(3^8\right)^{2020}\left(mod11\right)\)

\(\left(3^8\right)^{2020}=3^{8.404.5}=\left(3^5\right)^{3232}=\left(243\right)^{3232}\)

\(243\equiv1\left(mod11\right)\Rightarrow243^{3232}\equiv1\left(mod11\right)\)

\(\Rightarrow\left(14^8\right)^{2020}\equiv1\left(mod11\right)\)

\(\Rightarrow\left(14^8\right)^{2020}+10⋮11\)

\(242=2\cdot11^2\)

=>Số ước là (1+1)x(2+1)=6(ước)

SỐ lít dầu ăn còn lại là:

\(3\times\left(1-\dfrac{2}{5}\right)=3\times\dfrac{3}{5}=\dfrac{9}{5}\left(lít\right)\)