Question

[Lạc Linh Miêu]

a)  \(\sqrt{x+1}\).(x + 4) = (18 + x)\(\sqrt{6+x}\)- 3x - 40

b)  3 (\(\sqrt{x+9}\)- \(\sqrt{x+1}\)) = 4 - 4x

GIẢI PT HỘ MIK VS NHA <3 PT NÀO CX ĐC, CẢ 2 THÌ CÀNG TỐT ~ ^.^ ~

Answer 1

[Thắng Nguyễn]

a)\(\sqrt{x+1}\left(x+4\right)=\left(x+18\right)\sqrt{6+x}-3x-40\)

\(pt\Leftrightarrow\sqrt{x+1}\left(x+4\right)-14=\left(x+18\right)\sqrt{6+x}-63-3x-9\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)^2-196}{\sqrt{x+1}\left(x+4\right)+14}=\frac{\left(x+18\right)^2\left(x+6\right)-3969}{\left(x+18\right)\sqrt{6+x}+63}-3\left(x-3\right)\)

\(\Leftrightarrow\frac{x^3+9x^2+24x-180}{\sqrt{x+1}\left(x+4\right)+14}-\frac{x^3+42x^2+540x-2025}{\left(x+18\right)\sqrt{6+x}+63}+3\left(x-3\right)=0\)

\(\Leftrightarrow\frac{\left(x-3\right)\left(x^2+12x+60\right)}{\sqrt{x+1}\left(x+4\right)+14}-\frac{\left(x-3\right)\left(x^2+45x+675\right)}{\left(x+18\right)\sqrt{6+x}+63}+3\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{x^2+12x+60}{\sqrt{x+1}\left(x+4\right)+14}-\frac{x^2+45x+675}{\left(x+18\right)\sqrt{6+x}+63}+3\right)=0\)

Pt trong ngoặc to to kia vô nghiệm

Suy ra x=3

b)\(3\left(\sqrt{x+9}-\sqrt{x+1}\right)=4-4x\)

\(pt\Leftrightarrow\sqrt{x+9}-\sqrt{x+1}=\frac{4-4x}{3}\)

\(\Leftrightarrow2x+10-2\sqrt{\left(x+1\right)\left(x+9\right)}=\frac{16x^2-32x+16}{9}\)

\(\Leftrightarrow-2\sqrt{\left(x+1\right)\left(x+9\right)}=\frac{16x^2-32x+16}{9}-\left(2x+10\right)\)

\(\Leftrightarrow4\left(x+1\right)\left(x+9\right)=\frac{256x^4-1600x^3+132x^2+7400x+5476}{81}\)

\(\Leftrightarrow\frac{-64\left(x^2-5x-5\right)\left(4x^2-5x-8\right)}{81}=0\)

mỗi lần bình phương tự rút ra điều kiện mà khử nghiệm nhé :v

Answer 2

[Lạc Linh Miêu]

hi, ^.^, thanks bn nhìu nha >3