30-(x+1)-(16-x)=sqrt{(x+1)^2+(16-x)^2} eching

$$30 - ( x + 1 ) - ( 16 - x ) = \sqrt { ( x + 1 ) ^ { 2 } + ( 16 - x ) ^ { 2 } }$$

x uchun yechish

$$x=11$$

$$x=4$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(30-\left(x+1\right)-\left(16-x\right)\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

x+1 teskarisini topish uchun har birining teskarisini toping.

$$\left(30-x-1-\left(16-x\right)\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

29 olish uchun 30 dan 1 ni ayirish.

$$\left(29-x-\left(16-x\right)\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

16-x teskarisini topish uchun har birining teskarisini toping.

$$\left(29-x-16+x\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

13 olish uchun 29 dan 16 ni ayirish.

$$\left(13-x+x\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

0 ni olish uchun -x va x ni birlashtirish.

$$13^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

2 daraja ko‘rsatkichini 13 ga hisoblang va 169 ni qiymatni oling.

$$169=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$169=\left(\sqrt{x^{2}+2x+1+\left(16-x\right)^{2}}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(16-x\right)^{2} kengaytirilishi uchun ishlating.

$$169=\left(\sqrt{x^{2}+2x+1+256-32x+x^{2}}\right)^{2}$$

257 olish uchun 1 va 256'ni qo'shing.

$$169=\left(\sqrt{x^{2}+2x+257-32x+x^{2}}\right)^{2}$$

-30x ni olish uchun 2x va -32x ni birlashtirish.

$$169=\left(\sqrt{x^{2}-30x+257+x^{2}}\right)^{2}$$

2x^{2} ni olish uchun x^{2} va x^{2} ni birlashtirish.

$$169=\left(\sqrt{2x^{2}-30x+257}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{2x^{2}-30x+257} ga hisoblang va 2x^{2}-30x+257 ni qiymatni oling.

$$169=2x^{2}-30x+257$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2x^{2}-30x+257=169$$

Ikkala tarafdan 169 ni ayirish.

$$2x^{2}-30x+257-169=0$$

88 olish uchun 257 dan 169 ni ayirish.

$$2x^{2}-30x+88=0$$

Ikki tarafini 2 ga bo‘ling.

$$x^{2}-15x+44=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon x^{2}+ax+bx+44 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-15 ab=1\times 44=44$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 44-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-44 -2,-22 -4,-11$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-44=-45 -2-22=-24 -4-11=-15$$

Yechim – -15 yigʻindisini beruvchi juftlik.

$$a=-11 b=-4$$

x^{2}-15x+44 ni \left(x^{2}-11x\right)+\left(-4x+44\right) sifatida qaytadan yozish.

$$\left(x^{2}-11x\right)+\left(-4x+44\right)$$

Birinchi guruhda x ni va ikkinchi guruhda -4 ni faktordan chiqaring.

$$x\left(x-11\right)-4\left(x-11\right)$$

Distributiv funktsiyasidan foydalangan holda x-11 umumiy terminini chiqaring.

$$\left(x-11\right)\left(x-4\right)$$

Tenglamani yechish uchun x-11=0 va x-4=0 ni yeching.

$$x=11 x=4$$

30-\left(x+1\right)-\left(16-x\right)=\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}} tenglamasida x uchun 11 ni almashtiring.

$$30-\left(11+1\right)-\left(16-11\right)=\sqrt{\left(11+1\right)^{2}+\left(16-11\right)^{2}}$$

Qisqartirish. x=11 tenglamani qoniqtiradi.

$$13=13$$

30-\left(x+1\right)-\left(16-x\right)=\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}} tenglamasida x uchun 4 ni almashtiring.

$$30-\left(4+1\right)-\left(16-4\right)=\sqrt{\left(4+1\right)^{2}+\left(16-4\right)^{2}}$$

Qisqartirish. x=4 tenglamani qoniqtiradi.

$$13=13$$

-\left(x+1\right)-\left(16-x\right)+30=\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}} boʻyicha barcha yechimlar roʻyxati.

$$x=11 x=4$$