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\(\dfrac{2x-2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+1}\)
ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
a, Ta có :
\(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x+\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}-2}\)sử dụng tam thức bậc 2 khai triển biểu thức trên tử nhé
\(=\frac{\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\left(\sqrt{x}\right)^3-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)
b, Ta có : \(P=Q\)hay \(2\sqrt{x}+1=x-1\Leftrightarrow-x+2\sqrt{x}+2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-3=0\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
TH1 : \(\sqrt{x}=1+\sqrt{3}\Leftrightarrow x=\left(1+\sqrt{3}\right)^2=1+2\sqrt{3}+3=4+2\sqrt{3}\)
TH2 : \(\sqrt{x}=1-\sqrt{3}\Leftrightarrow x=\left(1-\sqrt{3}\right)^2=1-2\sqrt{3}+3=4-2\sqrt{3}\)
Vậy \(x=4+2\sqrt{3};x=4-2\sqrt{3}\)thì P = Q
んuﾘ ｲ giải pt vô tỉ không xét ĐK là tai hại :))
\(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\left(x\sqrt{x}-\sqrt{x}\right)+\left(2x-2\right)}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}=\frac{\left(x-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=x-1\)
Để P = Q thì \(2\sqrt{x}+1=x-1\)( x ≥ 1 ; x ≠ 4 )
<=> \(x-2\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-1\right)^2-3=0\)
<=> \(\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
<=> \(\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=4+2\sqrt{3}\left(tm\right)\\x=4-2\sqrt{3}\left(ktm\right)\end{cases}}\)
Vậy với \(x=4+2\sqrt{3}\)thì P = Q
2/ x2 + 2x - 2x - 9√x + 14 = ( x2 - 2x + 1) + (2x - 2×2×9√x /4 + 81/16) + 127/16 = (x - 1)2 + [ √(2x) - 9/4]2 + 127/16 > 0 với mọi x>= 1
Vậy phương trình vô nghiệm
Bài rút gọn để rút gọn được tử với mẫu thì phải phân tích được ra nhân tử chung cho cả tử và mẫu mà ta thấy tử không thể phân tích thành nhân tử được do tử luôn >0. Mẫu và tử lại cùng bậc nữa nên mình đầu hàng không rút gọn được
ĐKXĐ: \(x\ge0;x\ne3\)
\(B=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3\sqrt{x}-3}{x-9}\)
\(M=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}\right).\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(\frac{3x\sqrt{x}+2x}{2x\sqrt{x}+x+\sqrt{x}-1}\)
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
Q=\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)+\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6-x+4\sqrt{x}-4+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x\sqrt{x}-2x+2-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}-1}=\sqrt{x}+1\)
\(Q=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{x-3\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{x-3\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6-x+4\sqrt{x}-4+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\sqrt{x}+1\)
Lời giải:
\(\frac{2x-2\sqrt{x}+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=2+\frac{2}{x-\sqrt{x}}\)
\(\dfrac{2x-2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+1}\)