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
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 \(S=\left\{1;6\right\}\)
Câu c,d làm giống câu b
Câu e làm giống câu a
f. ĐKXĐ : \(x\ge1\)
Ta có :
\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\\ \Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1}=1\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{x-1}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-\sqrt{x-1}=1\) (*)
+) Nếu \(x\ge2\\ \Rightarrow x-1\ge1\\ \Rightarrow\sqrt{x-1}\ge1\\ \Rightarrow\sqrt{x-1}-1\ge0\)
=> (*) \(\Leftrightarrow\sqrt{x-1}-1-\sqrt{x-1}=1\)
\(\Leftrightarrow-1=1\) (vô lí)
+) Nếu \(x< 2\\ \Rightarrow x-1< 1\\ \Rightarrow\sqrt{x-1}-1< 0\)
=> (*) \(\Leftrightarrow1-\sqrt{x-1}-\sqrt{x-1}=1\\ \Leftrightarrow-2\sqrt{x-1}=0\\ \Leftrightarrow\sqrt{x-1}=0\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy \(S=\left\{1\right\}\)
Từ câu g đến câu m đưa về bình phương như câu f :
\(\sqrt{A^2}=\left|A\right|\)
Sau đó xét dấu giá trị tuyệt đối và giải pt như bình thường
\(a)\sqrt {x + 3} - \sqrt {x - 4} = 1 (\text{Điều kiện}: x \ge 0)\)
\( \Leftrightarrow \sqrt {x + 3} = 1 + \sqrt {x - 4} \\ \Leftrightarrow x + 3 = 1 + 2\sqrt {x - 4} + x - 4\\ \Leftrightarrow x + 3 = - 3 + 2\sqrt {x - 4} + x\\ \Leftrightarrow - 2\sqrt {x - 4} = - 3 - 3\\ \Leftrightarrow - 2\sqrt {x - 4} = - 6\\ \Leftrightarrow \sqrt {x - 4} = 3\\ \Leftrightarrow x - 4 = 9\\ \Leftrightarrow x = 9 + 4\\ \Leftrightarrow x = 13 \text{(thỏa mãn điều kiện)} \)
\(b)\sqrt {10 - x} + \sqrt {x + 3} = 5. \text{(Điều kiện}: -3 \le x \le 10) \\ \Leftrightarrow \sqrt {10 - x} = 5 - \sqrt {x + 3} \\ \Leftrightarrow 10 - x = 25 - 10\sqrt {x + 3} + x + 3\\ \Leftrightarrow 10 - x = 28 - 10\sqrt {x + 3} + x\\ \Leftrightarrow 10\sqrt {x + 3} = 2x + 18\\ \Leftrightarrow 5\sqrt {x + 3} = x + 9\\ \Leftrightarrow 25\left( {x + 3} \right) = 81 + 18x + {x^2}\\ \Leftrightarrow 25x + 75 - 81 - 18x - {x^2} = 0\\ \Leftrightarrow - {x^2} + 7x - 6 = 0\\ \Leftrightarrow {x^2} - 7x + 6 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1 \text{(thỏa mãn điều kiện)} \\ x = 6 \text{(thỏa mãn điều kiện)} \end{array} \right. \)
