v uchun yechish (complex solution)
v=\frac{\sqrt{2x+3}-\sqrt{x}}{\left(x+1\right)\left(x+3\right)}
x\neq -3\text{ and }x\neq -1
v uchun yechish
v=\frac{\sqrt{2x+3}-\sqrt{x}}{\left(x+1\right)\left(x+3\right)}
x\geq 0
Grafik
Baham ko'rish
Klipbordga nusxa olish
\sqrt{2x+3}-\sqrt{x}=\left(x+1\right)\left(x+3\right)v
Tenglamaning ikkala tarafini \left(x+1\right)\left(x+3\right) ga ko'paytirish.
\sqrt{2x+3}-\sqrt{x}=\left(x^{2}+4x+3\right)v
x+1 ga x+3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.
\sqrt{2x+3}-\sqrt{x}=x^{2}v+4xv+3v
x^{2}+4x+3 ga v ni ko'paytirish orqali distributiv xususiyatdan foydalanish.
x^{2}v+4xv+3v=\sqrt{2x+3}-\sqrt{x}
Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.
\left(x^{2}+4x+3\right)v=\sqrt{2x+3}-\sqrt{x}
v'ga ega bo'lgan barcha shartlarni birlashtirish.
\frac{\left(x^{2}+4x+3\right)v}{x^{2}+4x+3}=\frac{\sqrt{2x+3}-\sqrt{x}}{x^{2}+4x+3}
Ikki tarafini x^{2}+4x+3 ga bo‘ling.
v=\frac{\sqrt{2x+3}-\sqrt{x}}{x^{2}+4x+3}
x^{2}+4x+3 ga bo'lish x^{2}+4x+3 ga ko'paytirishni bekor qiladi.
v=\frac{\sqrt{2x+3}-\sqrt{x}}{\left(x+1\right)\left(x+3\right)}
\sqrt{2x+3}-\sqrt{x} ni x^{2}+4x+3 ga bo'lish.
\sqrt{2x+3}-\sqrt{x}=\left(x+1\right)\left(x+3\right)v
Tenglamaning ikkala tarafini \left(x+1\right)\left(x+3\right) ga ko'paytirish.
\sqrt{2x+3}-\sqrt{x}=\left(x^{2}+4x+3\right)v
x+1 ga x+3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.
\sqrt{2x+3}-\sqrt{x}=x^{2}v+4xv+3v
x^{2}+4x+3 ga v ni ko'paytirish orqali distributiv xususiyatdan foydalanish.
x^{2}v+4xv+3v=\sqrt{2x+3}-\sqrt{x}
Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.
\left(x^{2}+4x+3\right)v=\sqrt{2x+3}-\sqrt{x}
v'ga ega bo'lgan barcha shartlarni birlashtirish.
\frac{\left(x^{2}+4x+3\right)v}{x^{2}+4x+3}=\frac{\sqrt{2x+3}-\sqrt{x}}{x^{2}+4x+3}
Ikki tarafini x^{2}+4x+3 ga bo‘ling.
v=\frac{\sqrt{2x+3}-\sqrt{x}}{x^{2}+4x+3}
x^{2}+4x+3 ga bo'lish x^{2}+4x+3 ga ko'paytirishni bekor qiladi.
v=\frac{\sqrt{2x+3}-\sqrt{x}}{\left(x+1\right)\left(x+3\right)}
\sqrt{2x+3}-\sqrt{x} ni x^{2}+4x+3 ga bo'lish.
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