giai phuong trinh \(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
giai phuong trinh sau:
\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}+\frac{1}{4}}=2\)
m.m giup minh cau nay aj. thank m.n
giai phuong trinh
\(18x^2-2x-\frac{17}{3}+9\sqrt{x-\frac{1}{3}}=0\)
\(18x^2-2x-\frac{17}{3}+9\sqrt{x-\frac{1}{3}}=0\)
Điều kiện: \(x\ge\frac{1}{3}\)
Đặt \(\sqrt{x-\frac{1}{3}}=a\left(a\ge0\right)\)
\(\Rightarrow x=a^2+\frac{1}{3}\)
Ta suy ra phương trình tương đương với
\(18\left(a^2+\frac{1}{3}\right)^2-2\left(a^2+\frac{1}{3}\right)-\frac{17}{3}+9a=0\)
\(\Leftrightarrow54a^4+30a^2+27a-13=0\)
\(\Leftrightarrow\left(3a-1\right)\left(18a^3+6a^2+12a+13\right)=0\)
Dễ thấy \(18a^3+6a^2+12a+13>0\) vì \(a\ge0\)
\(\Rightarrow3a-1=0\)
\(\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{x-\frac{1}{3}}=\frac{1}{3}\)
\(\Leftrightarrow x-\frac{1}{3}=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{4}{9}\)
giai phuong trinh : \(\frac{1}{1+x}+\frac{1}{1+\sqrt{x}}=\frac{2+\sqrt{x}}{2x}\)
2.giai phuong trinh sau:
a.\(\sqrt{\frac{2x-3}{x-1}}=2\)
b.\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
Đkiện: x <1 hoặc x \(\ge\frac{3}{2}\)
\(\sqrt{\frac{2x-3}{x-1}}=2\) (1)
(1) => \(\frac{2x-3}{x-1}=4\)
=> 2x - 3 = 4x - 4
<=> 2x - 4x = -4 + 3
<=> -2x = -1
<=> x = \(\frac{1}{2}\)( TMĐK)
Vậy x = \(\frac{1}{2}\)
b, Đkiện: x \(\ge\frac{3}{2}\)
(1) => \(\sqrt{2x-3}=2\sqrt{x-1}\)
=>2x - 3 = 4(x - 1)
<=> 2x -3 = 4x -4
<=> -2x = -1
<=> x = \(\frac{1}{2}\)(ko TMĐK)
Vậy pt vô nghiệm
b. \(x>0;x\ne1\)
\(\Rightarrow\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\Rightarrow2x-3=4x-4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
giai phuong trinh: \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x-1}\)
giai phuong trinh
\(\frac{1}{\sqrt{x}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}=1\)
\(DK:x\ge0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)
\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)
\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow x=1\)
Vay nghiem cua PT la \(x=1\)
giup minh bai nay:
giai phuong trinh:
x(x+1)(x-1)(x+2)=24
2x(8x-1)^2(3x+2)(2x+1)=3
thanks!
a) [x(x+1].[(x-1)(x+2)]=24
(x2+x)(x2+x+2)=24
Dat x2+x=a , ta dc: a(a+2)=24
=> a2+2a-24=0
=> (a-4)(a+6)=0
=> a=4 hoac a=-6
Thay vao roi tu tim x nha
b)
Giai phuong trinh :\(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x+8}=1+\sqrt{3}\)
