bài 1

\(\frac{x-1}{x+3}>0\)   \(\left(x\ne-3\right)\)

   TH1  \(\hept{\begin{cases}x-1>0\\x+3< 0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x>1\\x< -3\end{cases}}\)(vô lí)

      TH2 \(\hept{\begin{cases}x-1< 0\\x+3>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x< 1\\x>-3\end{cases}}\)\(\Rightarrow-3< x< 1\)

bài 2 . với dạng này ta áp dụng bđt \(|x|< A\Leftrightarrow\orbr{\begin{cases}x< -A\\x>A\end{cases}}\)

|x - 5| >2

\(\Leftrightarrow\orbr{\begin{cases}x-5>2\\x-5< -2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x>7\\x< 3\end{cases}}\)

#mã mã#

a) \(\Delta'=m^2-\left(m-4\right)=m^2-m+4=m^2-2.m.\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\)

\(=\left(m-\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0;\forall m\)

=> phương trình (1) luôn có hai nghiệm phân biệt với mọi m

b) Áp dụng định lí Viet ta có: 

\(x_1.x_2=m-4\)

\(x_1+x_2=-2m\)

=> \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1.x_2=\left(-2m\right)^2-2\left(m-4\right)=4m^2-2m+8\)

=> \(x_1^3+x_2^3=\left(x_1+x_2\right)\left(x_1^2-x_1x_2+x_2^2\right)=\left(-2m\right)\left(4m^2-2m+8-\left(m-4\right)\right)\)

\(=-2m\left(4m^2-3m+12\right)\)

Theo bài ra ta có:

 \(x_1+x_2=\frac{x_1^2}{x_2}+\frac{x_2^2}{x_1}\)

 \(\Leftrightarrow x_1+x_2=\frac{x_1^3+x_2^3}{x_1.x_2}\)

Thay vào ta có:

\(-2m=\frac{-2m\left(4m^2-3m+12\right)}{m-4}\)( đk m khác 4)

\(\Leftrightarrow\orbr{\begin{cases}m=0\\m-4=4m^2-3m+12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}m=0\left(tm\right)\\4m^2-4m+16=0\left(l\right)\end{cases}\Leftrightarrow m=0}\)

Vì \(4m^2-4m+16=\left(2m-1\right)^2+15>0\) với mọi m

Vậy m =0

cảm ơn nhìu

a) Từ I hạ IH,IK lần lượt vuông góc với AB,AC. Theo tính chất điểm nằm trên phân giác của góc thì IH = IK.

Xét \(\Delta\)IHE và \(\Delta\)IKD: IH = IK, ^IHE = ^IKD = 90⁰, IE = ID (gt) => \(\Delta\)IHE = \(\Delta\)IKD (Ch.cgv)

=> ^IEH = ^IDK hay ^IEA = ^IDC => Tứ giác ADIE nội tiếp

=> ^BAC = 180⁰ - ^DIE = 180⁰ - ^BIC = 180⁰ - (180⁰ - ^ABC/2 - ^ACB/2) = ^ABC/2 + ^ACB/2

= 90⁰ - ^BAC/2 => 3.^BAC = 180⁰ => ^BAC = 60⁰. Vậy góc BAC = 60⁰.

b) Trên cạnh BC lấy điểm F sao cho IF là phân giác của ^BIC.

Theo câu a: ^BAC = 60⁰, tứ giác ADIE nội tiếp => ^DIE = ^BIC = 120⁰ => ^BIF = ^CIF = 60⁰

Mà ^BIE = ^CID = ^BAC = 60⁰ nên ^BIE = ^BIF = ^CIF = ^CID

Dễ dàng chỉ ra \(\Delta\)BEI = \(\Delta\)BFI (g.c.g), \(\Delta\)CDI = \(\Delta\)CFI (g.c.g)

=> BE = BF,CD = CF. Do đó BE + CD = BC. Tức là \(\frac{BE}{BC}+\frac{CD}{BC}=1\)

Áp dụng ĐL đường phân giác trong tam giác: \(\frac{BE}{BC}=\frac{AE}{AC}\left(=\frac{IE}{IC}\right)=\frac{BE+AE}{BC+AC}=\frac{AB}{BC+AC}\)

Từ đó \(\frac{AB}{BC+CA}+\frac{AC}{AB+BC}=1\)=> \(\frac{AB+BC+CA}{AB+BC}+\frac{AB+BC+CA}{BC+CA}=3\)

Vậy thì \(\frac{1}{AB+BC}+\frac{1}{BC+CA}=\frac{3}{AB+BC+CA}\) (đpcm).

hình như là -4

\(\left|x-3\right|=\left|x+5\right|\)

\(\Rightarrow\orbr{\begin{cases}x-3=x+5\\x-3+x+5=0\end{cases}}\)

\(\Rightarrow2x+2=0\)

=> 2x = -2

=> x = -1

vậy_

\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)

\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)

\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)

\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)

\(S=\frac{\left(2019-1\right)}{2}=1009\)

\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)

\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)

\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)

\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)

• a,de bieu thuc 2x^2+3x-1 /2x+1 nguyen thi2x+1 phai khac 0suy ra 2x khac -1suy ra x khac -1/2b,de bieu thuc 2x^2+4x-1/2x+1 nguyen thi 2x+1 khac 0suy ra 2x khac -1suy ra x khac -1/2c,de bieu thuc 2x^2+3/x-2 nguyen thi x-2 khac 0suy ra x khac 2 

\(a,=\frac{\left(2x+1\right)\left(x+1\right)-2}{2x+1}\in Z\Leftrightarrow2⋮2x+1mà:2x+1le\Rightarrow2x+1\in\left\{1;-1\right\}\Leftrightarrow x\in\left\{-1;0\right\}\)

ĐK: \(x^2-1\ge0\) (1)

\(pt\Leftrightarrow\left(x^2+3\sqrt{x^2-1}\right)^2=x^4-x^2+1\)

\(\Leftrightarrow6\sqrt{x^2-1}+9\left(x^2-1\right)=-x^2+1\)

\(\Leftrightarrow6\sqrt{x^2-1}+10\left(x^2-1\right)=0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(6+10\sqrt{x^2-1}\right)=0\)

\(\Leftrightarrow\sqrt{x^2-1}=0\)

\(\Leftrightarrow x^2-1=0\Leftrightarrow x=\pm1\)Thỏa mãn đk (1)

Vậy ...

\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)

\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)

\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\

\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=10\)

\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)

\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)

\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)

\(=\sqrt{3}-1\)