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\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
a) 3x - 2 = 0 => 3x = 2 => x = 2/3
b) 2x - 1 = 0 => 2x = 1 => x = 1/2
c) 5 ( 4+2x) = 8+5x
<=> 20 + 10x = 8 + 5x
<=> 10x - 5x = 8 - 20
<=> 5x = -12
x = -12/5
d) \(\frac{1}{2}+\frac{3}{4}x=6-\frac{4}{5}x\)
\(\frac{3}{4}x+\frac{4}{5}x=6-\frac{1}{2}\)
\(\frac{31}{20}x=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{31}{20}=\frac{110}{31}\)
e) 3 + 2x = 4 - 8x
<=> 2x + 8x = 4 - 3
10 x = 1
x = 1/10
f \(5+\frac{1}{2}\left(x+5\right)=3\)
\(\frac{1}{2}\left(x+5\right)=3-5=-2\)
\(x+5=-2:\frac{1}{2}=-4\)
\(x=-4-5=1\)
Vậy ......
Câu a:
- \(\frac12\)(3 - 2\(x\)) - 7 = 5 - \(\frac13\)(\(x\) - \(\frac45\))
- \(\frac32\) + \(x\) - 7 = 5 - \(\frac13x\) + \(\frac{4}{15}\)
\(x\) + \(\frac13x\) = 5 + \(\frac{4}{15}\) + 7 + \(\frac32\)
(1 + \(\frac13\))\(x\) = \(\frac{150}{30}\) + \(\frac{8}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{158}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{368}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{413}{30}\)
\(x\) = \(\frac{413}{30}\) : \(\frac43\)
\(x\) = \(\frac{413}{40}\)
Vậy \(x=\frac{413}{40}\)
Câu b:
(5 - 3x/2) : - 1 3/8 = - 7 1/3
(5 - 3x/2) : (-11/8) = - 22/3
5 - 3x/2 = - 22/3 x (-11/8)
5 - 3x/2 = 121/12
3x/2 = 5 - 121/12
3x/2 = - 61/12
x = - 61/12 : 3/2
x = -61/18
Vậy x = - 61/18
Bài 1:
a) Chỗ y6 là 6.y hay là y6
b) \(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Rightarrow2x-2-6x-6-8x-12=16\)
\(\Rightarrow\left(2x-6x-8x\right)-\left(2+6+12\right)=16\)
\(\Rightarrow-12x-20=16\)
\(\Rightarrow-12x=36\)
\(\Rightarrow x=-3\)
Vậy x = -3
c) \(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+13}=0\)
\(\Rightarrow\left(x-5\right)^{x+1}\left[1-\left(x-5\right)^{12}\right]=0\)
\(\Rightarrow\left(x-5\right)^{x+1}=0\) hoặc \(1-\left(x-5\right)^{12}=0\)
+) \(\left(x-5\right)^{x+1}=0\Rightarrow x-5=0\Rightarrow x=5\)
+) \(1-\left(x-5\right)^{12}=0\Rightarrow\left(x-5\right)^{12}=1\)
\(\Rightarrow x-5=\pm1\)
+) \(x-5=1\Rightarrow x=6\)
+) \(x-5=-1\Rightarrow x=4\)
Vậy \(x\in\left\{6;4\right\}\)
Bài 2: a, thiếu dữ liệu
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\left[\begin{matrix}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{a}=1\end{matrix}\right.\Rightarrow\left[\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)
Ta có: \(\frac{a^3b^2c^{1930}}{a^{1935}}=\frac{a^3a^2a^{1930}}{a^{1935}}=\frac{a^{1935}}{a^{1935}}=1\)
Vậy \(\frac{a^3b^2c^{1930}}{a^{1935}}=1\)
sửa câu a bài 1 là y6 là bỏ 6 đi
câu 2a) 2x^2 + 2y^2 -3z^2 = -100