đề j v

Phần a dễ bạn tự làm nha!!! :))

b, Ta có: \(\Delta^'=\left[-\left(m+1\right)\right]^2-2m=m^2+2m+1-2m=m^2+1>0\forall m\)

=> PT luôn có 2 nghiệm phân biệt

Theo Vi-ét, ta có: \(\hept{\begin{cases}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{cases}}\)

Ta có: \(\sqrt{x_1}+\sqrt{x_2}=\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=2\)

\(\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=2\)

\(\Leftrightarrow x_1+x_2-2+2\sqrt{x_1x_2}=0\)

\(\Leftrightarrow2\left(m+1\right)-2+2\sqrt{2m}=0\)

\(\Leftrightarrow2m+2\sqrt{2m}=0\)

\(\Leftrightarrow m+\sqrt{2m}=0\)

\(\Leftrightarrow\sqrt{m}\left(\sqrt{m}+\sqrt{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{m}=0\\\sqrt{m}+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\\sqrt{m}=-\sqrt{2}\end{cases}}}\)

Vậy: m = 0

=.= hk tốt!!

a) Khi m=1 thì pt<=>x²-4x+2=0

Có:\(\Delta\)'=(-2)^2-2=2>0=>pt có 2 nghiệm là x₁=\(2+\sqrt{2}\)và x₂=2-\(\sqrt{2}\)

b)Để pt có nghiệm thì \(\Delta\)'=(m+1)²-2\(\ge\)0<=>m\(\ge\)\(\sqrt{2}\)-1

Theo định lý Viète thì:x₁+x₂=2(m+1)=\(\sqrt{2}\)<=>\(\frac{\sqrt{2}-2}{2}\)

Tham khảo:

Chúc bn học tốt

u​i a lê​n gp nhanh v

\(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\sqrt{\left(x-1\right)^2}+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|x-6\right|\right)\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\)

Ta có \(\hept{\begin{cases}\left|x-4\right|\ge0\forall x\\\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\end{cases}}\)

=> \(\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\ge5\forall x\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-4=0\\\left(x-1\right)\left(6-x\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

=> MinA = 5 <=> x = 4

Ta có: \(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(\Rightarrow A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left|x-1\right|+\left|x-6\right|\)

Xét \(\left|x-1\right|+\left|x-6\right|\)ta có: 

\(\left|x-1\right|+\left|x-6\right|=\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\)(1)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(6-x\right)\ge0\)

TH1: Nếu \(\hept{\begin{cases}x-1< 0\\6-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\6< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>6\end{cases}}\)( vô lý )

TH2: Nếu \(\hept{\begin{cases}x-1\ge0\\6-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\6\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le6\end{cases}}\Leftrightarrow1\le x\le6\)

mà \(\left|x-4\right|\ge0\)(2)

Từ (1) và (2) \(\Rightarrow A\ge5\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-4=0\\1\le x\le6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

Vậy \(minA=5\)\(\Leftrightarrow x=4\)

1: ĐKXĐ: (4-x)(x+2)>=0

=>(x-4)(x+2)<=0

=>-2<=x<=4

Ta có: \(x^2-2x-8=4\sqrt{\left(4-x\right)\left(x+2\right)}\)

=>\(4\sqrt{-\left(x-4\right)\left(x+2\right)}=x^2-2x-8\)

=>\(4\sqrt{-\left(x^2-2x-8\right)}=x^2-2x-8\)

=>\(\begin{cases}x^2-2x-8\ge0\\ -16\left(x^2-2x-8\right)=\left(x^2-2x-8\right)^2\end{cases}\Rightarrow\begin{cases}\left(x-4\right)\left(x+2\right)\ge0\\ \left(x^2-2x-8\right)\left(x^2-2x-8+16\right)=0\end{cases}\)

=>\(\begin{cases}\left(x-4\right)\left(x+2\right)\ge0\\ \left(x-4\right)\left(x+2\right)\left(x^2-2x+8\right)=0\end{cases}\Rightarrow\begin{cases}\left(x-4\right)\left(x+2\right)\ge0\\ \left(x-4\right)\left(x+2\right)=0\end{cases}\)

=>(x-4)(x+2)=0

=>x=-2(nhận) hoặc x=4(nhận)

Ta có; \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{x^2-2x+1}{2}\)

\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\left(x-1\right)^2}{2}\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)

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

a/ ĐKXĐ: \(1\le x\le5\)

- Với \(4< x\le5\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng

- Với \(x\le4\) hai vế ko âm, bình phương:

\(-x^2+6x-5>64-32x+4x^2\)

\(\Leftrightarrow5x^2-38x+69< 0\) \(\Rightarrow3< x< \frac{23}{5}\)

Vậy nghiệm của BPT là: \(3< x\le5\)

b/ ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge-\frac{4}{3}\end{matrix}\right.\)

- Với \(x< 1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm

- Với \(x\ge1\) hai vế ko âm, bình phương:

\(\left(x+5\right)\left(3x+4\right)< 16\left(x-1\right)^2\)

\(\Leftrightarrow13x^2-51x-4>0\)

\(\Rightarrow x>4\)