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b) Theo bài ra ta có : \(2x+3y-z=50\)

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2\left(x-1\right)}{4}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{2x-2}{4}=\dfrac{3y-2}{9}=\dfrac{z-3}{4}=\\ \dfrac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\ =\dfrac{2x-2+3y-6-z+3}{9}\\ =\dfrac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}\\ =\dfrac{50-5}{9}\\ =\dfrac{45}{9}\\ =5\\ \)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow2x=22\Rightarrow x=11\\\dfrac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow3y=51\Rightarrow y=17\\\dfrac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\end{matrix}\right.\)

\(\text{Vậy }x=11\\ y=17\\ z=23\)

a) Theo bài ra ta có : \(x+y+z=49\)

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\\ =\dfrac{12x+12y+12z}{18+16+15}\\ =\dfrac{12\left(x+y+z\right)}{49}\\ =\dfrac{12\cdot49}{49}\\ =12\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x}{18}=12\Rightarrow12x=216\Rightarrow x=18\\\dfrac{12y}{16}=12\Rightarrow12y=192\Rightarrow y=16\\\dfrac{12z}{15}=12\Rightarrow12z=180\Rightarrow z=15\end{matrix}\right.\)

\(\text{Vậy }x=18\\ y=16\\ z=15\)

\(\text{a) }0.3\left(12\right)=\dfrac{312-3}{990}=\dfrac{103}{330}\)

\(\text{b) }2.0\left(33\right)=2\dfrac{033-0}{990}=2\dfrac{33}{990}=3\dfrac{1}{30}=\dfrac{91}{30}\)

Đứng thú 81 thế là thế nào

\(\dfrac{9^4\cdot2^{15}}{2^6\cdot24^3}=\dfrac{\left(3^2\right)^4\cdot2^{15}}{2^6\cdot\left(8\cdot3\right)^3}=\dfrac{3^8\cdot2^{15}}{2^6\cdot8^3\cdot3^3}=\dfrac{3^8\cdot2^{15}}{2^6\cdot2^9\cdot3^3}=\dfrac{3^8\cdot2^{15}}{2^{15}\cdot3^3}=3^5=243\)

Chứng minh bằng đồng vị

Tính \(M_2=135^0\)

Mà \(M_2=M_4\left(\text{2 góc đối đỉnh}\right)\)

\(\Rightarrow M_4=135^0\)

Mà \(P_1=135^0\)

\(P_1=M_4\)

Mà \(P_1\) và \(M_4\) lại là 2 góc đồng vị

\(\Rightarrow a\text{//}b\left(ĐPCM\right)\)

Vậy \(a\text{//}b\)

Chứng minh bằng trong cùng phía :

Chứng minh \(M_4=135^0\left(1\right)\)

Ta có : \(P_1+P_2=180^0\left(\text{2 góc kề bù}\right)\)

Hay : \(135^0+P_2=180^0\)

\(\Rightarrow P_2=45^0\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra :

\(M_4+P_2=180^0\left(135^0+45^0=180^0\right)\)

Mà \(M_4\) và \(P_2\) là 2 góc trong cùng phía

\(\Rightarrow a\text{//}b\left(ĐPCM\right)\)

Vậy \(a\text{//}b\)

\(A=x^2+x+1\\ A=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\\ A=x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\\ A=\left(x+\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\\ A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(\text{Mà }\left(x+\dfrac{1}{2}\right)^2\ge0\\ \Rightarrow A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\left(ĐPCM\right)\)

Vậy \(A\ge\dfrac{3}{4}\)

b trong duoc so cay gap hau lan si cay trong cua xa a la sao sai den roi

Choose the letter A, B, C or D the word that has the underlined part different from others.

A. funny

B. uncle

C. curious

D. fund