1:

\(A=\left\{0;1;2;4\right\};B=\left\{0;2;3;5;7\right\}\)

\(A\cap B=\left\{0;2\right\}\)

\(A\cup B=\left\{0;1;2;4;3;5;7\right\}\)

\(A\text{B}=\left\{1;4\right\}\)

\(B\text{A}=\left\{3;5;7\right\}\)

2: \(A\text{B}=\left\{1;4\right\}\)

\(A=\left\{0;1;2;4\right\}\)

=>(A\B)\(\subset\)A

A\(A\B)

={0;1;2;4}\{1;4}

={0;2}

=\(A\cap B\)

1)

a) \(0,7\left(6\right)=\dfrac{76-7}{90}=\dfrac{69}{90}=\dfrac{23}{30}\)

b) ta có: \(0,2\left(148\right)=\dfrac{2148-2}{9990}=\dfrac{2146}{9990}=\dfrac{29}{135}\)

do đó: \(1,2\left(148\right)=1+0,2\left(148\right)=1+\dfrac{29}{135}=\dfrac{164}{135}\)

cảm ơn ạ!!

Lời giải:

a) Ta có:

$x^2+2x+3=0$

$\Leftrightarrow (x^2+2x+1)=-2$

$\Leftrightarrow (x+1)^2=-2< 0$ (vô lý do $(x+1)^2\geq 0, \forall x$)

Do đó PT vô nghiệm

b)

$(x+3)^2-6x=0$

$\Leftrightarrow x^2+6x+9-6x=0$

$\Leftrightarrow x^2=-9< 0$ (vô lý)

Do đó PT vô nghiệm.

\(\begin{array}{l}C_5^0 - C_5^1 + C_5^2 - C_5^3 + C_5^4 - C_5^5\\ = C_5^0{.1^5} - C_5^1{.1^4}.1 + C_5^2{.1^3}{.1^2} - C_5^3{.1^2}{.1^3} + C_5^4{.1.1^4} - C_5^5{.1^5}\\ = {\left( {1 - 1} \right)^5} = {0^5}\\ = 0\end{array}\)

Vậy ta có điều phải chứng minh

Cách 2:

Ta có: \(C_5^0 = C_5^{5 - 0} = C_5^5\)

Tương tự: \(C_5^1 = C_5^{5 - 1} = C_5^4;\;C_5^2 = C_5^{5 - 2} = C_5^3;\)

\(\Rightarrow C_5^0 - C_5^1 + C_5^2 - C_5^3 + C_5^4 - C_5^5 = \left( {C_5^0 - C_5^5} \right) + \left( {C_5^4 - C_5^1} \right) + \left( {C_5^2 - C_5^3} \right) = 0\) (đpcm)

a)

\(\begin{array}{l}C_4^0 + 2C_4^1 + {2^2}C_4^2 + {2^3}C_4^3 + {2^4}C_4^4\\ = {1^4}.C_4^0 + {1^3}.2C_4^1 + {1^2}{.2^2}C_4^2 + {1.2^3}C_4^3 + {2^4}C_4^4\\ = {\left( {1 + 2} \right)^4} = {3^4}\end{array}\)

\( = 81\) (đpcm)

b)

\(\begin{array}{l}C_4^0 - 2C_4^1 + {2^2}C_4^2 - {2^3}C_4^3 + {2^4}C_4^4\\ = {1^4}.C_4^0 - {1^3}.2C_4^1 + {1^2}{.2^2}C_4^2 - {1.2^3}C_4^3 + {2^4}C_4^4\\ = {\left( {1 - 2} \right)^4} = {\left( { - 1} \right)^4}\end{array}\)

\( = 1\) (đpcm)