Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Giải phương trình :
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\) .
Bài Toán :
Giải phương trình sau :
\(\frac{3\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\frac{4.\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{5\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)
Bài 1: Rút gọn
a. \(\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2\)
b. \(\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}\)
c. \(\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}\)
d. \(\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}\)
e. \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}\)
Bài 2: Rút gọn
A =\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{\sqrt{x+1}}{x-2\sqrt{x}+1}\right)\)(x>0 ; x khác 1)
a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:
1) \(x^2-3x-3=\frac{3\left(\sqrt[3]{x^3-4x^2+4}-1\right)}{1-x}\) ;2)\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
b) Giải các phương trình sau(không giới hạn phương pháp):
1)\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\) ; 2)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
3)\(\frac{3x^2+3x-1}{3x+1}=\sqrt{x^2+2x-1}\) ; 4) \(\frac{2x^3+3x^2+11x-8}{3x^2+4x+1}=\sqrt{\frac{10x-8}{x+1}}\)
5)\(13x-17+4\sqrt{x+1}=6\sqrt{x-2}\left(1+2\sqrt{x+1}\right)\);
6)\(x^2+8x+2\left(x+1\right)\sqrt{x+6}=6\sqrt{x+1}\left(\sqrt{x+6}+1\right)+9\)
7)\(x^2+9x+2+4\left(x+1\right)\sqrt{x+4}=\frac{5}{2}\sqrt{x+1}\left(2+\sqrt{x+4}\right)\)
8)\(8x^2-26x-2+5\sqrt{2x^4+5x^3+2x^2+7}\)
1)\(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
2)\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)
3)\(9+3\sqrt{x\left(3-2x\right)}=7\sqrt{x}+5\sqrt{3-2x}\)
4)\(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
5)\(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)
6)\(2\left(5x-3\right)\sqrt{x+1}+\left(x+1\right)\sqrt{3-x}=3\left(5x+1\right)\)
7)\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
Giải phương trình:
a) \(\left|x+1\right|+\left|x-1\right|=1+\left|x^2-1\right|\)
b) \(\frac{x^2}{3}+\frac{48}{x^2}=10\left(\frac{x}{3}-\frac{4}{x}\right)\)
c) \(2\sqrt[3]{x^2}-5\sqrt[3]{x}=3\)
a) Cho x = \(\frac{\sqrt[3]{10+6\sqrt{3}}\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)Tính giá trị biểu thức: A = \(\left(x^3-4x+1\right)^{2018}\)
b) Cho x = \(\sqrt[3]{7+5\sqrt{2}}-\frac{1}{\sqrt[3]{7+5\sqrt{2}}}\)Tính giá trị biểu thức: B = \(\left(x^3+3x-14\right)^{2018}\)
Giải giúp mik vs đap cần gấp . Cảm ơn mn. Giải cho mik bài 1 cx đc
1/ Rút gọn
A=\(\sqrt{7}-4\sqrt{3}+\sqrt{4}-2\sqrt{3}\)
B=\(\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right)\) \(\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
C=\(\left(\sqrt{3}+1\right)\) \(\left(\frac{\sqrt{14}-6\sqrt{3}}{5+\sqrt{3}}\right)\)
2/Cho P=\(\left(\sqrt{x-\frac{1}{\sqrt{x}}}\right)\):\(\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
a/ cmr: P>0, V x >0, x\(\ne\)1
b/Tính GT P khi x\(\frac{2}{2+\sqrt{3}}\)
1/Tính:
a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
b)\(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right).\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
2/Giải phương trình:
a)\(\sqrt{x^2-2x+1}=7\)
b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)