ĐKXĐ: \(x\le2\)\(\Rightarrow\sqrt{2-x}\ge0\)
Ta có : \(\sqrt[3]{2x^2+6x+3}=\sqrt[3]{2\left(x+\dfrac{3}{2}\right)^2-\dfrac{3}{2}}\ge\sqrt[3]{-\dfrac{3}{2}}\)
\(\Rightarrow\sqrt{2-x}+\sqrt[3]{2x^2+6x+3}\ge\sqrt[3]{-\dfrac{3}{2}}\)
mặt khác \(-2=\sqrt[3]{\dfrac{-16}{2}}< \sqrt[3]{\dfrac{-3}{2}}\)
\(\Rightarrow VT>VP\)
vậy phương trình vô nghiệm
gpt :A= \(2x^2-5x-1=\sqrt{x+2}+\sqrt{4-x}\)
B= \(\sqrt{x^2-2x+5}+2\sqrt{4x+5}=x^3-2x^2+5x+4\)
gpt a/ \(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)
b/ \(2\sqrt{1-x}-\sqrt{1+x}+3\sqrt{1-x^2}=3-x\)
Gpt
\(\sqrt{3x+3}-\sqrt{5-2x}-x^3+3x^2+10x-16=0\)
gpt:
\(\sqrt{7x^2+25x+19}-\sqrt{x^2-2x-35}=7\sqrt{x+2}\)
Các bạn giúp mình với. Tks
1/ Gpt 4x2 - 8(2x+3)\(\sqrt{2x-1}=7-44x\)
2/ gpt x3- x2 - 10x -2 = \(\sqrt[3]{7x^2+23x+12}\)
3/ Choa,b,c > 0 và thỏa a+b+c =3
CM : \(\frac{a}{a+2bc}+\frac{b}{b+2ca}+\frac{c}{c+2ab}\ge1\)
gpt ; a, 3\(\sqrt{x^2+x+1}\) - x = x2 +3
b, \((\) 2x2 + x - 2\()\)2 + 10x2 +5x - 16 = 0
1, \(x^3-x-3=2\sqrt{6x-x^2}\)
2, \(x^3+6x^2-171x-40\left(x+1\right)\sqrt{5x-1}+20=0\)
3, \(\sqrt[3]{x+3}+\sqrt[3]{x-3}=\sqrt[5]{x-5}+\sqrt[5]{x+5}\)
4. \(\left(\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{x+1}\right)^2=\frac{4\left(1+\sqrt{1+4x}\right)}{x+1+\sqrt{x^2+3x+2}}\)
Giải các phương trình sau:
a) \(2x-x^2+\sqrt{6x^2-12x+7}\)
b) (x+1)(x+4)-3\(\sqrt{x^2+5x+2}\)
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
