Question

[Nguyễn Thị Hằng]

Giải các phương trình sau:

1.

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

b. \(\sqrt{10-x}+\sqrt{x+3}=5\)

c. \(\sqrt{15-x}+\sqrt{3-x}=6\)

d. \(\sqrt{x-1}+\sqrt{x+1}=2\)

e. \(\sqrt{4x+1}-\sqrt{3x+4}=1\)

f. \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

g. \(\sqrt{x+\sqrt{2x+1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)

h. \(\sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt{6}\)

i. \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)

k. \(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+9-6\sqrt{x}}=1\)

l. \(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)

m. \(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}=1}\)

n. \(\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1\)

o. \(\sqrt{1-\sqrt{x^2-x}}=\sqrt{x}-1\)

p. \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)

q. \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

r. \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)

s. \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)

t. \(\sqrt{3x+15}-\sqrt{4x-17}=\sqrt{x+2}\)

u. \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}=4-2x\)

v. \(\sqrt{x+1}+\sqrt{x+10}=\sqrt{x+2}+\sqrt{x+5}\)

w. \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)

x. \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)

y. \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\dfrac{x-1}{x-2}}=3\)

z. \(\left(x-2\right)\left(x+2\right)+4\left(x-2\right)\sqrt{\dfrac{x+2}{x-2}}=-3\)

2.

a. \(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)

b. \(\dfrac{x}{2+\dfrac{x}{2+\dfrac{x}{2+\dfrac{...}{2+\dfrac{x}{1+\sqrt{1+x}}}}}}=8\) (vế trái có 100 dấu phân thức)

c. \(\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)

d. \(\sqrt[4]{1-x}+\sqrt[4]{2-x}=\sqrt[4]{3-2x}\)

e. \(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}+\sqrt[4]{1-x}=3\)

f. \(\dfrac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)

g. \(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0\)

h. \(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{\left(x-1\right)^2}+\sqrt[3]{x^2-1}=1\)

i. \(\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}\)

k. \(\sqrt[3]{x-2}+\sqrt{x+1}=3\)

l. \(\sqrt[3]{24+x}+\sqrt{12-x}=6\)

m. \(\sqrt[3]{2-x}+\sqrt{x-1}=1\)

n. \(1+\sqrt[3]{x-16}=\sqrt[3]{x+3}\)

o. \(\sqrt[3]{25+x}+\sqrt[3]{3-x}=4\)

p. \(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)

Làm nhanh giúp mk nhé mn ơi

Answer 1

[Thiện Đạt Hoàng Nghĩa]

Giải pt :

1

a. ĐKXĐ : \(x\ge4\)

Ta có :

\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)

\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)

\(\Leftrightarrow x=13\) (TM ĐKXĐ)

Vậy \(S=\left\{13\right\}\)

b.ĐKXĐ : \(-3\le x\le10\)

Ta có :

\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy \(S=\left\{1;6\right\}\)

Answer 2

[Thiện Đạt Hoàng Nghĩa]

Câu c,d làm giống câu b

Câu e làm giống câu a