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Lag tí -.-'
`ĐK:2<=x<=6`
BP 2 vế ta có:
`x-2+6-x+2\sqrt{(x-2)(6-x)}=x^2-8x+24`
`<=>4+2\sqrt{(x-2)(6-x)}=x^2-8x+24`
`<=>2\sqrt{(x-2)(6-x)}=x^2-8x+20`
`<=>2sqrt{-x^2+8x-12}=x^2-8x+20`
`<=>-x^2+8x-20+2sqrt{-x^2+8x-12}=0`
`<=>-x^2+8x-12+2sqrt{-x^2+8x-12}-8=0`
Đặt `sqrt{-x^2+8x-12}=a(a>=0)`
`pt<=>a^2+2a-8=0`
`<=>a=2(tm),a=-4(l)`
`<=>-x^2+8x-12=4`
`<=>x^2-8x+16=0`
`<=>(x-4)^2=0<=>x=4(tmđk)`
Vậy `S={4}`
\(\sqrt{x-2}+\sqrt{6-x}\text{=}\sqrt{x^2-8x+24}\)
\(ĐKXĐ:2\le x\le6\)
Xét VP của pt ta thấy : \(\sqrt{x^2-8x+24}\text{=}\sqrt{x^2-8x+16+8}\)
\(\text{=}\sqrt{\left(x-4\right)^2+8}\)
\(\Rightarrow VP\ge\sqrt{8}\)
Xét VT của pt ta có :
\(VT^2\text{=}x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)
\(VT^2\text{=}4+2\sqrt{\left(x-2\right)\left(6-x\right)}\)
Áp dụng BĐT cô si cho 2 số không âm ta có :
\(2\sqrt{\left(x-2\right)\left(6-x\right)}\le\left(\sqrt{x-2}\right)^2+\left(\sqrt{6-x}\right)^2\)
\(\text{=}x-2+6-x\text{=}4\)
\(\Rightarrow VT^2\le8\)
\(\Rightarrow VT\le\sqrt{8}\)
Để \(VT\text{=}VP\) \(\Leftrightarrow\left\{{}\begin{matrix}x-4\text{=}0\\\sqrt{x-2}\text{=}\sqrt{6-x}\end{matrix}\right.\)
\(\Leftrightarrow x=4\left(TM\right)\)
Vậy...........
Ghi thiếu đề bài nên tl lại
`sqrt{x-2}+sqrt{6-x}=x^2-8x+16+2sqrt2`
Áp dụng BĐT bunhia ta có:
`sqrt{x-2}+sqrt{6-x}<=sqrt{(1+1)(x-2+6-x)}=2sqrt2`
`=>VT<=2sqrt2(1)`
Mặt khác:
`VP=x^2-8x+16+2sqrt2`
`=(x-4)^2+2sqrt2>=2sqrt2`
`=>VP>=2sqrt2(2)`
`(1)(2)=>VT=VP=2sqrt2`
`<=>x=4`
Vậy `S={4}`
`sqrt{x-2}+sqrt{6-x}=x^2-8x+2sqrt2`
Áp dụng BĐT bunhia ta có:
`sqrt{x-2}+sqrt{6-x}<=sqrt{(1+1)(x-2+6-x)}=2sqrt2`
`=>VT<=2sqrt2(1)`
Mặt khác:
`VP=x^2-8x+16+2sqrt2`
`=(x-4)^2+2sqrt2>=2sqrt2`
`=>VP>=2sqrt2(2)`
`(1)(2)=>VT=VP=2sqrt2`
`<=>x=4`
Vậy `S={4}`
Áp dụng bđt Bunhia,ta có VT^2<=2(x-2+6-x)=8
suy ra VT<=\(2\sqrt{2}\)
Dấu "=" xảy ra khi \(\sqrt{x-2}=\sqrt{6-x}\) <=> x-2=6-x <=>x=4
Mặc khác \(\sqrt{x^2-8x+24}=\sqrt{\left(x-4\right)^2+8}>=2\sqrt{2}\)
Dấu "=" xảy ra khi \(\left(x-4\right)^2\)=0 <=> x=4
Vậy pt đã cho có 1 nghiệm duy nhất là x=4
ĐKXĐ: \(x\ge0\)
\(\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow...\)
Đk: `1 <=x <=7`.
Đặt `sqrt(7-x) = a, sqrt(x-1) = b`.
Phương trình trở thành: `b^2+1 + 2a = 2b + ab + 1`.
`<=> b^2 + 2a = 2b + ab.`
`<=> b(b-2) = a(b-2)`
`<=> (b-a)(b-2) = 0`
`<=> a =b` hoặc `b = 2.`
`@ a = b => 7 - x = x - 1`
`<=> 8 = 2x <=> x = 4`.
`@ b = 2 => sqrt(x-1) = 2`
`<=> x - 1 = 4`
`<=> x = 5`.
Vậy `x = 4` hoặc `x = 5`.
\(\text{ĐKXĐ:}1\le x\le7\)
PT đã cho tương đương với:
\(x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{x-1}.\sqrt{7-x}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{4;5\right\}\)
a) ĐK: \(0\le x\le\frac{\sqrt{5}+1}{2}\)
\(\sqrt{1-\sqrt{x^2-x}}=\sqrt{x}-1\)
\(\Leftrightarrow1-\sqrt{x^2-x}=\left(\sqrt{x}-1\right)^2\left(x\ge1\right)\)
\(\Leftrightarrow1-\sqrt{x^2-x}=x-2\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x\left(x-1\right)}=2\sqrt{x}-x\)
\(\Leftrightarrow\sqrt{x\left(x-1\right)}=\sqrt{x}\left(2-\sqrt{x}\right)\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-1}+\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x-1}+\sqrt{x}-2=0\end{cases}}\)
TH1: x = 0 (Loại)
TH2: \(\sqrt{x-1}+\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x-1}=2-\sqrt{x}\)
\(\Leftrightarrow x-1=4-4\sqrt{x}+x\left(x\le4\right)\)
\(\Leftrightarrow4\sqrt{x}=5\Leftrightarrow\sqrt{x}=\frac{5}{4}\Leftrightarrow x=\frac{25}{16}\left(tm\right)\)
b) \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
ĐK: \(x\ge1\)
\(pt\Leftrightarrow\sqrt{\left(x+1\right)\left(2x+6\right)}+\sqrt{\left(x+1\right)\left(x-1\right)}=2\left(x+1\right)\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{2x+6}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{2x+6}+\sqrt{x-1}-2\sqrt{x+1}=0\end{cases}}\)
TH1: \(\sqrt{x+1}=0\Leftrightarrow x=-1\left(l\right)\)
TH2: \(\sqrt{2x+6}=2\sqrt{x+1}-\sqrt{x-1}\)
\(\Leftrightarrow2x+6=4\left(x+1\right)+\left(x-1\right)-4\sqrt{x^2-1}\)
\(\Leftrightarrow2x+6=5x+3-4\sqrt{x^2-1}\)
\(\Leftrightarrow4\sqrt{x^2-1}=3x-3\Leftrightarrow16\left(x^2-1\right)=9x^2-18x+9\left(x\ge1\right)\)
\(\Leftrightarrow7x^2+18x-25=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=-\frac{25}{7}\left(l\right)\end{cases}}\)
dk tu xd \(\sqrt{2x^2+8x+6}\) \(+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}-\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\)
\(\Leftrightarrow\sqrt{x+1}\left(2\sqrt{x+3}-\sqrt{x-1}-2\sqrt{x+1}\right)=0\)
đến đây bn tự giải nhé
ĐKXĐ: \(2\le x\le6\)
\(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\\ \Leftrightarrow\left(\sqrt{x-2}+\sqrt{6-x}\right)^2=\left(\sqrt{x^2-8x+24}\right)^2\\ \Leftrightarrow x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\\ \Leftrightarrow4+2\sqrt{-x^2+8x-12}=x^2-8x+24\\ \Leftrightarrow-x^2+8x-20+2\sqrt{-x^2+8x-12}=0\left(1\right)\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\), ta có:
\(\left(1\right)\Leftrightarrow a^2+2a-8=0\Leftrightarrow\left[{}\begin{matrix}a=2\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)
Ta có:
\(\sqrt{-x^2+8x-12}=2\Leftrightarrow-x^2+8x-12=4\\ \Leftrightarrow-x^2+8x-16=0\\ \Leftrightarrow x^2-8x+16=0\\ \Leftrightarrow\left(x-4\right)^2=0\\ \Leftrightarrow x=4\left(tm\right)\)
Vậy....
P.s: Có gì sai mong mọi người góp ý!
#Lemon
ĐK:....
\(pt\Leftrightarrow x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow4+2\sqrt{-x^2+8x-12}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(x^2-8x=a\)
\(pt\Leftrightarrow2\sqrt{-a-12}=a+20\)
\(\Leftrightarrow4\left(-a-12\right)=\left(a+20\right)^2\)
\(\Leftrightarrow a^2+40a+400+4a+48=0\)
\(\Leftrightarrow a^2+44a+448=0\)
\(\Leftrightarrow\left(a+16\right)\left(a+28\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-16\\a=-28\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+16=0\\x^2-8x+28=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\\left(x-4\right)^2+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\varnothing\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất \(x=4\)