Chương I - Căn bậc hai. Căn bậc ba

đề bài là gì vậy?

\(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}=1\)

<=> \(\sqrt{\left(\sqrt{x-2}\right)^2+2\sqrt{x-2}+1}+\sqrt{\left(\sqrt{x-2}\right)^2-2\sqrt{x-2}+1}=1\)

<=> \(\sqrt{\left(\sqrt{x-2}+1\right)^2}+\sqrt{\left(\sqrt{x-2}-1\right)^2}=1\)

<=> \(|\sqrt{x-2}+1|+|\sqrt{x-2}-1|=1\)

<=> \(\sqrt{x-2}+1+\sqrt{x-2}-1=1\)

<=> \(2\)\(\sqrt{x-2}=1\)

<=> \(\sqrt{x-2}=\dfrac{1}{2}\)

<=> x - 2 = \(\dfrac{1}{4}\)

<=> x = \(\dfrac{1}{4}+2\)

<=> x = \(\dfrac{9}{4}\)

Vậy...............

Sai thì thôi nha......mik mới hok cái này trong sgk nên chưa chắc

câu 1

1)

a)

\(\sqrt{\sqrt{17}+1}\cdot\sqrt{\sqrt{17}-1}=\sqrt{\left(\sqrt{17}+1\right)\left(\sqrt{17}-1\right)}=\sqrt{17-1}=\sqrt{16}=4\)b)

\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}=\left|2-\sqrt{3}\right|+\sqrt{3}=2-\sqrt{3}+\sqrt{3}\text{(vì 2>\sqrt{3})}=2\)c)

\(\dfrac{2}{\sqrt{3}-1}-\sqrt{3}\left(\sqrt{12}+1\right)=\dfrac{2}{\sqrt{3}-1}-6-\sqrt{3}=\dfrac{2-6\sqrt{3}+6-3+\sqrt{3}}{\sqrt{3}-1}=\dfrac{5-5\sqrt{3}}{\sqrt{3}-1}=-5\)2)

ta có pt hoành độ giao điểm

\(\left(m-3\right)x+5=2x-m+1\)

mà 2 đường thẳng trên giao nhau trên trục tung nên có hoành độ bằng 0

\(\Rightarrow5=-m+1\Leftrightarrow m=-4\)

vậy m=-4

3)

a) đặt 2y-5=x ta có

\(x^2-x-12=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2y-5=4\\2y-5=-3\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}y=\dfrac{9}{2}\\y=1\end{matrix}\right.\)b)

\(\left\{{}\begin{matrix}3x-y=5\\x+2y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6x-2y=10\\x+2y=4\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)mình chỉ giúp được tớ đây thôi mình hết kiên nhẫn rồi nên khi nào rảnh mình giải cho

Ta có \(ab+bc+ca=2abc\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=2\\P=\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{y^3}{\left(2-y\right)^3}+\dfrac{z^3}{\left(2-z\right)^2}\end{matrix}\right.\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{2-x}{8}+\dfrac{2-x}{8}\ge3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(2-y\right)^2}+\dfrac{2-y}{8}+\dfrac{2-y}{8}\ge\dfrac{3y}{4}\\\dfrac{z^3}{\left(2-z\right)^2}+\dfrac{2-z}{8}+\dfrac{2-z}{8}\ge\dfrac{3z}{8}\end{matrix}\right.\)

\(\Rightarrow P+\dfrac{12-2\left(x+y+z\right)}{8}\ge\dfrac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\dfrac{1}{2}\)

Dấu " = " xảy ra khi \(x=y=z=\dfrac{2}{3}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{1-a}{8}+\dfrac{1-a}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{1-b}{8}+\dfrac{1-b}{8}\ge\dfrac{3b}{4}\\\dfrac{c^3}{\left(1-c\right)^2}+\dfrac{1-c}{8}+\dfrac{1-c}{8}\ge\dfrac{3c}{4}\end{matrix}\right.\)

\(\Rightarrow P+\dfrac{6-2\left(a+b+c\right)}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{1}{4}\)

Vậy \(P_{min}=\dfrac{1}{4}\)

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)

a

bạn có chơi fb ko

Giải:

Xét hiệu \(A=\left(a^2+b^2+1\right)-\left(ab+a+b\right)\)

\(=a^2+b^2+1-ab-a-b\)

\(\Rightarrow2A=2a^2+2b^2+2-2ab-2a-2b\)

\(=\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)\) \(+\left(b^2-2b+1\right)\)

\(=\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)

\(\Rightarrow2A\ge0\Leftrightarrow A\ge0\)

Vậy \(a^2+b^2+1\ge ab+a+b\) (Đpcm)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)

ta có: \(A=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)

\(\Leftrightarrow A^2=\left(\sqrt{10}-\sqrt{2}\right)^2\left(3+\sqrt{5}\right)\)

\(\Leftrightarrow A^2=\left(12-4\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(\Leftrightarrow A^2=36+12\sqrt{5}-12\sqrt{5}-20\)

\(\Leftrightarrow A^2=16\)

\(\Leftrightarrow A=4\)

Vậy A=4

Đặt t bằng x²

Đk t >= 0

Ta có phương trình

t² - 8t + t +12 = 0

đen ta = (-8)²-1*(t+12)

= 64-t-12

=52 - t

Suy ra 52 - t > 0\(\Leftrightarrow\) -t > -52\(\Leftrightarrow\) t < 52

t₁₌ --8 - \(\sqrt{52}\)/ 2*1=4+\(\sqrt{13}\) (nhận)

t₂₌ --8 +\(\sqrt{52}\)/ 2*1=4-\(\sqrt{13}\) (nhận )

đợt nọ my teacher làm như thế này:

vì \(x\ge-1\)nên \(\sqrt{x+1}\ge0\)

\(\sqrt{x+10}\ge3\)

\(VT\ge3\)

tương tự \(VF\ge3\)

nên VT=VT <=> x=-1

p/s:I don't sure about that