hướng dẫn giải

B1 : bn chia nó ra làm hai bước tính trong một phép .

vd : 3x/8 = 3y/64

tương tự như vậy

còn cách tính làm sao thì dễ rồi nha

\(\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{64}=\dfrac{z}{216}\)

đặt x/8=y/64=z/216=k

=>x=8k; y=64k; z=216k

\(2x^2+2y^2-z^2=1\)

\(\Leftrightarrow128k^2+2\cdot64^2\cdot k^2-\left(216k\right)^2=1\)

\(\Leftrightarrow k^2=\dfrac{1}{-38336}\)(vô lý)

\(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{210}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{64}=\dfrac{z}{210}\\ \Rightarrow\dfrac{x^2}{64}=\dfrac{y^2}{4096}=\dfrac{z^2}{44100}\\ \Rightarrow\dfrac{x^2}{64}=\dfrac{y^2}{4096}=\dfrac{z^2}{44100}=\dfrac{2x^2+2y^2-z^2}{2.64+2.4096-44100}=\dfrac{1}{-35780}\\ \Rightarrow x;y;z\in\varnothing\)

a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x

Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y

= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y

Ta lại có : 1 + 4y/24 = 1+4y / 9+3y

=> 24=9+3y => 15=3y => y=5

Vậy y=5

Nhớ like

b, 1+3y/12 = 1+5y/5x = 1+7y/4x

Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x

= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x

Ta lại có: 1+5y / 5x = 1+5y / 6+2x

=> 5x = 6+2x => 3x = 6 => x=2

Vậy x =2

Mình sửa lại câu a

1+2y/18 = 1+6y / 6x = 1+2y+1+6y / 18 + 6x = 2 +8y /18+6x

= 2 (1+4y) / 2 (9 +3x) = 1+4y / 9 +3x

Ta lại có: 1+4y/24 = 1+4y/ 9 +3x

=> 24 = 9 +3x => 15= 3x => x =5

2x²+2y²+2z²=1 hay (2x)²+(2y)²+(2z)²=1

\(\text{Ta có:}\)\(\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}\)

\(\Rightarrow\frac{3}{8}x=\frac{3}{8}.\frac{y}{8}=\frac{3}{8}.\frac{z}{17}\)

\(\Rightarrow x=\frac{y}{8}=\frac{z}{27}\)

\(\text{Đặt:}\)\(x=\frac{y}{8}=\frac{z}{27}=k\)

\(\Rightarrow x=k\)

\(\frac{y}{8}=k\Rightarrow y=8k\)

\(\frac{z}{27}=k\Rightarrow z=27k\)

\(\text{Có:}\)\(2x^2-2y^2-z^2=1\)

\(\Rightarrow\left(2k\right)^2+2\left(8k^2\right)-27k^2=1\)

\(\Rightarrow k^2.\left(2+2.8^2-27^2\right)=1\)

\(\Rightarrow k^2.\left(-599\right)=1\)

\(\Rightarrow k^2=\frac{-1}{599}\left(\text{Vô lí}\right)\)

\(\Rightarrow x,y,z\text{ ko có gtrị}\)