ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある代数 >

n(x-3)=sqrt((x-3)-3)

解

n (x - 3) = (x - 3) - 3

解

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}

解答ステップ

n (x - 3) = (x - 3) - 3

拡張

n (x - 3) : n x - 3 n

n x - 3 n = x - 3 - 3

両辺を2乗する:

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 6

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 6

解く

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 6 : x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}}

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}}

解を検算する:

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}

解答は

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}

グラフ

人気の例

sqrt(x-\sqrt{2x)-1}=0

x - 2 x - 1 = 0

1=(\sqrt[3]{x+2})/4

1 = \frac{3 x + 2}{4}

sqrt(a)-sqrt(5)=sqrt(10)

a - 5 = 10

3 x^{2} + 2 = 0

sqrt(7x+5)=sqrt(x-3)

7 x + 5 = x - 3