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Đặt \(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=B\) = B
Xét tích \(AB=\left(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}\right)\left(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15}\right)\)
\(=x^2-7x+19-\left(x^2-7x+15\right)=x^2-7x+19-x^2+7x-15\)
\(=4\)
Mà \(B=2\Leftrightarrow A=2\)
ta có:\(\frac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}=\frac{1}{2}\)
\(\Rightarrow x-3\sqrt{x}+1=0\)
\(\Rightarrow\hept{\begin{cases}x+1=3\sqrt{x}\\x-3\sqrt{x}=-1\end{cases}}\)
lại có \(B=\frac{3x\sqrt{x}+10x+19}{x^2+7x+15}\)
\(=\frac{3x\sqrt{x}-9x+19x+19}{x^2-9x+16x+15}\)
\(=\frac{3\sqrt{x}\left(x-3\sqrt{x}\right)+19\left(x+1\right)}{\left(x+3\sqrt{x}\right)\left(x-3\sqrt{x}\right)+16x+15}\)
\(=\frac{-3\sqrt{x}+19\times3\sqrt{x}}{-1\times\left(x+3\sqrt{x}\right)+16x+15}\)
\(=\frac{57\sqrt{x}-3\sqrt{x}}{15x+15-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{15\left(x+1\right)-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{45\sqrt{x}-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{42\sqrt{x}}=\frac{27}{21}\)
Ta có:
\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}\)
\(=\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{75}{4}}+\sqrt{\left(2x-1\right)^2+3\left(x+2\right)^2}+\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{3}{4}\left(4x+3\right)^2}\)
\(\ge\sqrt{\frac{75}{4}}+\sqrt{3\left(x+2\right)^2}+\sqrt{\frac{3}{4}\left(4x+3\right)^2}\)
\(=\frac{5\sqrt{3}}{2}+\sqrt{3}\left(x+2\right)+\frac{\sqrt{3}\left(4x+3\right)}{2}=3\sqrt{3}\left(x+2\right)\)
Dấu = xảy ra khi ....
ĐK:\(3x^2-6x-6\ge0;\left(2-x\right)^5\ge0;x\le2\)
\(\Leftrightarrow\sqrt{3x^2-6x-6}-\sqrt{3}=3\sqrt{\left(2-x\right)^5}-27\sqrt{3}+\left(7x-19\right)\sqrt{2-x}+26\sqrt{3}\)
\(\Leftrightarrow\frac{3x^2-6x-9}{\sqrt{3x^2-6x-6}+\sqrt{3}}=3\left(\frac{\left(2-x\right)^5-243}{\sqrt{\left(2-x\right)^5}+9\sqrt{3}}\right)+\frac{\left(7x-19\right)^2\left(2-x\right)-2028}{\left(7x-19\right)\sqrt{2-x}-26\sqrt{3}}\)
\(\Leftrightarrow\left(x+1\right)\left[...\right]=0\)
Ta c/m đc [...] khác 0.
Vậy x=-1(TM)
ĐKXĐ: \(x\le1-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3x^2-6x-6}=3\left(2-x\right)^2\sqrt{2-x}+\left(7x-19\right)\sqrt{2-x}\)
\(\Leftrightarrow\sqrt{3x^2-6x-6}=\left(3x^2-5x-7\right)\sqrt{2-x}\)
\(\Leftrightarrow\sqrt{3x^2-6x-6}=\left(3x^2-6x-6-\left(2-x\right)+1\right)\sqrt{2-x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x^2-6x-6}=a\ge0\\\sqrt{2-x}=b\ge0\end{matrix}\right.\) ta được:
\(a=\left(a^2-b^2+1\right)b\Leftrightarrow b\left(a^2-b^2\right)+b-a=0\)
\(\Leftrightarrow b\left(a+b\right)\left(a-b\right)-\left(a-b\right)=0\)
\(\Leftrightarrow\left(ab+b^2-1\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=1-b^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x^2-6x-6}=\sqrt{2-x}\\\sqrt{\left(3x^2-6x-6\right)\left(2-x\right)}=1-\left(2-x\right)\end{matrix}\right.\)
TH1: \(3x^2-5x-8=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{8}{3}>1-\sqrt{3}\left(l\right)\end{matrix}\right.\)
TH2: \(\sqrt{\left(3x^2-6x-6\right)\left(2-x\right)}=x-1\)
Do \(x\le1-\sqrt{3}\Rightarrow x-1\le-\sqrt{3}\Rightarrow VP< 0\) và \(VT\ge0\Rightarrow ptvn\)
Vậy pt có nghiệm duy nhất \(x=-1\)
Bài 2:
\(B=\sqrt{28-16\sqrt{3}}+\sqrt{13-4\sqrt{3}}\)
\(=\sqrt{\left(4-2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=\left|4-2\sqrt{3}\right|+\left|2\sqrt{3}-1\right|\)
\(=4-2\sqrt{3}+2\sqrt{3}-1\)
\(=3\)
\(C=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{2}.\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=5-3=2\)
\(D=\sqrt{4+2\sqrt{3}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{3}+1-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
1) \(\Leftrightarrow x^2-7x+8+\sqrt{x^2-7x+8}-20=0\)
Đặt \(t=\sqrt{x^2-7x+8}\ge0\)
Phương trình tương đương
\(t^2+t-20=0\)
\(\left[{}\begin{matrix}t=4\left(TM\right)\\t=-5\left(KTM\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-7x+8}=4\)
Bạn đọc tự giải quyết tiếp bài toán.
Hãy tích cho tui đi
vì câu này dễ mặc dù tui ko biết làm
Yên tâm khi bạn tích cho tui
Tui sẽ ko tích lại bạn đâu
THANKS
( x +1 ) ( x + 4 ) = 5 căn ( x^2 + 5x +28 ) (1)
= ( x + 1 ) ( x + 4 ) = 5 căn [ (x^2 + 5x + 4) + 24 ]
= ( x + 1 ) ( x + 4 ) = 5 căn [ ( x + 1 ) ( x + 4 ) + 24 ]
Đặt a = ( x + 1 ) ( x + 4 )
(1) <=> a = 5 căn ( a + 24 )
<=> a^2 = 25 ( a + 24 )
<=> a^2 - 25a - 600 = 0
<=> a1 = 40
a2 = -15
với a = 40 ta có:
( x + 1 ) ( x + 4 ) = 40
<=> x^2 + 5x + 4 = 40
<=> x^2 + 5x - 36 = 0
<=> x = 4 và x = - 9
với a = -15, ta có:
( x + 1 ) ( x + 4 ) = -15
<=> x^2 + 5x + 4 = -15
<=> x^2 + 5x + 19 = 0
delta < 0 => pt vô nghiệm
Vậy s = { -9; 4}
Ta có $\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=2$
$=>2M=(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15})(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15})$
$=>2M=\sqrt{x^2-7x+19}^2-\sqrt{x^2-7x+15}^2$
$=>2M=(x^2-7x+19)-(x^2-7x+15)=4$
$=>M=2$
\(2.M=\left(x^2-7x+19\right)-\left(x^2-7x+15\right)=4\Rightarrow M=2\)