Dat \(\sqrt{2x+1}=a,\sqrt{x}=b=>x=b^2\)
=> a.b=\(\sqrt{2x^2+x}\)
pt<=> 3(a+2b2+11)=4ab
=> 3a+6b2+33=4ab
Dat \(\sqrt{2x+1}=a,\sqrt{x}=b=>x=b^2\)
=> a.b=\(\sqrt{2x^2+x}\)
pt<=> 3(a+2b2+11)=4ab
=> 3a+6b2+33=4ab
a)\(\sqrt{4x-1}+\sqrt{4x^2-1}=1\)
b)\(x+4\sqrt{x+3}+2\sqrt{3-2x}=11\)
c)\(\sqrt{x}-x=1-\sqrt{2x+1}\)
d)\(\sqrt{x}+\sqrt{4-x}-2=-x\)
e)\(\sqrt{4+x}+x=\sqrt{4+12x}\)
Phương trình ✔2x-3=1 tương đương vs pt nào dưới đây ?
A.x✔2x-3=x
B.✔x-3+✔2x-3=1+✔x-3
C.(x-4)✔2x-3=x-4
D.(3-x)✔2x-3=3-x
Mong mn giúp với ạ
Giai pt
1,\(\sqrt{x+8-6\sqrt{x-1}}\)=4
2,\(\sqrt{x+6-2\sqrt{x+2}}\)+\(\sqrt{x+11-6\sqrt{x+2}}\)=1
3,\(\sqrt{x-3-2\sqrt{x-4}}\)+\(\sqrt{x-4\sqrt{x-4}}\)=1
4,\(\sqrt{x-2+\sqrt{2x+5}}\)+\(\sqrt{x+2+3\sqrt{2x-5}}\)=\(\dfrac{7}{2}\)
5,\(\sqrt{2x+4+6\sqrt{2x-5}}\)+\(\sqrt{2x-4-2\sqrt{2x-5}}\)=4
6,\(\sqrt{\dfrac{1}{4}x^2+x+1}\)-\(\sqrt{6-2\sqrt{5}}\)=0
7,x+\(\sqrt{x+\dfrac{1}{2}}\)+\(\sqrt{x+\dfrac{1}{4}}\)=2
8,\(\sqrt{\left(x-1\right)+4-4\sqrt{x-1}}+\sqrt{x-1-6\sqrt{x-1+9}}\)=1
9,\(\sqrt{x+2\sqrt{x-1}}\)+\(\sqrt{x-2\sqrt{x-1}}\)=\(\dfrac{x+3}{2}\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Căn x+(√2x-1) + căn x+4-3√2x-1 =4√2
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)
1.\(\sqrt{\frac{\left(1-x\right)}{x}}=\frac{\left(2x+x^2\right)}{1+x^2}\)
2. 3(2-\(\sqrt{x+2}\))=2x+\(\sqrt{x+6}\)
3. \(\sqrt[3]{x+2}+\sqrt[3]{x+1}=\sqrt[2]{2x^2}+\sqrt[3]{2x^2+1}\)
4. \(\sqrt[3]{x+24}+\sqrt{12-x}=6\)
Toán 10 ạ, giúp em với
Nghiệm của phương trình \(x^4+2x^3+2x^2-2x+1=\left(x^3+x\right)\sqrt{\frac{1}{x}-x}\) có dạng \(a+\sqrt{b}\) với a, b thuộc Z. Tính ab.
gpt :A= \(2x^2-5x-1=\sqrt{x+2}+\sqrt{4-x}\)
B= \(\sqrt{x^2-2x+5}+2\sqrt{4x+5}=x^3-2x^2+5x+4\)