2: \(-4x^2+5x-2\)

\(=-4\left(x^2-\dfrac{5}{4}x+\dfrac{1}{2}\right)\)

\(=-4\left(x^2-2\cdot x\cdot\dfrac{5}{8}+\dfrac{25}{64}+\dfrac{7}{64}\right)\)

\(=-4\left(x-\dfrac{5}{8}\right)^2-\dfrac{7}{16}< =-\dfrac{7}{16}< 0\forall x\)

Sửa đề:\(f\left(x\right)=\dfrac{-x^2+4\left(m+1\right)x+1-4m^2}{-4x^2+5x-2}\)

Để f(x)>0 với mọi x thì \(\dfrac{-x^2+4\left(m+1\right)x+1-4m^2}{-4x^2+5x-2}>0\forall x\)

=>\(-x^2+4\left(m+1\right)x+1-4m^2< 0\forall x\)(1)

\(\text{Δ}=\left[\left(4m+4\right)\right]^2-4\cdot\left(-1\right)\left(1-4m^2\right)\)

\(=16m^2+32m+16+4\left(1-4m^2\right)\)

\(=32m+20\)

Để BĐT(1) luôn đúng với mọi x thì \(\left\{{}\begin{matrix}\text{Δ}< 0\\a< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}32m+20< 0\\-1< 0\left(đúng\right)\end{matrix}\right.\)

=>32m+20<0

=>32m<-20

=>\(m< -\dfrac{5}{8}\)

a: ĐKXĐ: x=0; x<>1

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

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

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

1.\(x=4\)

\(B=\left(\dfrac{x+1}{2}-\sqrt{x}\right)=\left(\dfrac{4+1}{2}-\sqrt{4}\right)=\dfrac{5}{2}--2=\dfrac{5-4}{2}=\dfrac{1}{2}\)

2.\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)=\left(\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

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

\(B=\dfrac{x+1}{2}-\sqrt{x}=\dfrac{x+1-2\sqrt{x}}{2}=\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

\(M=A.B=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

3.\(M=\dfrac{\sqrt{x}}{6}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{6}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=6\left(\sqrt{x}-1\right)\)

\(\Leftrightarrow x+\sqrt{x}=6\sqrt{x}-6\)

\(\Leftrightarrow x-5\sqrt{x}+6=0\)

Đặt \(\sqrt{x}=a;a\ge0\)

=> pt trở thành:

\(a^2-5a+6=0\)

\(\Delta=\left(-5\right)^2-4.6=25=24=1>0\)

=> pt có 2 nghiệm:

\(\left\{{}\begin{matrix}x_1=\dfrac{5+\sqrt{1}}{2}=3\left(tm\right)\\x_2=\dfrac{5-\sqrt{1}}{2}=2\left(tm\right)\end{matrix}\right.\)

Xét \(\sqrt{a}=3\)

\(\Leftrightarrow a=9\)

Xét \(\sqrt{a}=2\)

\(\Leftrightarrow a=4\)

Vậy \(x=9;4\)

`(2m-10)^2/((m+5)^2+1)`

`=(2m-10)^2/(m^2+10m+26)-404+404`

`=(4m^2-40m+100)/(m^2+10m+26)-404+404`

`=(4m^2-40m+100-404m^2-4040m-10504)/(404[(m+5)^2+1])+404`

`=(-400m^2-4080m-10404)/(404[(m+5)^2+1])+404`

`=(-(400m^2+4080m+10404))/(404[(m+5)^2+1])+404`

`=(-(20m+102)^2)/(404[(m+5)^2+1])+404<=404`

Dấu "=" xảy ra khi `20m+102=0<=>m=(-51)/10`

Bài này giải kiểu lớp 8 thì nó cực kì vô duyên:

\(P=\dfrac{4m^2-40m+100}{m^2+10m+26}=\dfrac{404\left(m^2+10m+26\right)-4\left(100m^2+1020m+2601\right)}{m^2+10m+26}\)

\(P=404-\dfrac{4\left(10m+51\right)^2}{\left(m+5\right)^2+1}\le404\)

\(P_{max}=404\) khi \(m=-\dfrac{51}{10}\)

câu 1) ta có : \(M=\left(x^2-x\right)^2+\left(2x-1\right)^2=x^4-2x^3+x^2+4x^2-4x+1\)

\(=\left(x^2-x+2\right)^2-3=\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right)^2-3\)

\(\Rightarrow\dfrac{1}{16}\le M\le61\)

\(\Rightarrow M_{min}=\dfrac{1}{16}\)khi \(x=\dfrac{1}{2}\) ; \(M_{max}=61\) khi \(x=3\)

câu 2) điều kiện xác định : \(0\le x\le2\)
đặt \(\sqrt{2x-x^2}=t\left(t\ge0\right)\)

\(\Rightarrow M=-t^2+4t+3=-\left(t-2\right)^2+7\)

\(\Rightarrow3\le M\le7\)

câu 3) ta có : \(M=\left(x-2\right)^2+6\left|x-2\right|-6\ge-6\)

\(\Rightarrow M_{min}=-6\) khi \(x=2\)

4) điều kiện xác định \(-6\le x\le10\)

ta có : \(M=5\sqrt{x+6}+2\sqrt{10-x}-2\)

áp dụng bunhiacopxki dạng căn ta có :

\(-\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\le5\sqrt{x+6}+2\sqrt{10-x}\le\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\)

\(\Leftrightarrow-4\sqrt{29}\le5\sqrt{x+6}+2\sqrt{10-x}\le4\sqrt{29}\)

\(\Rightarrow-2-4\sqrt{29}\le B\le-2+4\sqrt{29}\)

\(\Rightarrow M_{max}=-2+4\sqrt{29}\) khi \(\dfrac{\sqrt{x+6}}{5}=\dfrac{\sqrt{10-x}}{2}\Leftrightarrow x=\dfrac{226}{29}\)

1.

\(A=2.\left|3x-1\right|-4\)

Ta có:

\(\left|3x-1\right|\ge0\) \(\forall x.\)

\(\Rightarrow2.\left|3x-1\right|\ge0\) \(\forall x.\)

\(\Rightarrow2.\left|3x-1\right|-4\ge-4\) \(\forall x.\)

\(\Rightarrow A\ge-4.\)

Dấu '' = '' xảy ra khi:

\(3x-1=0\)

\(\Rightarrow3x=0+1\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\frac{1}{3}.\)

Vậy \(MIN_A=-4\) khi \(x=\frac{1}{3}.\)

2.

\(B=10-4.\left|x-2\right|\)

Ta có:

\(\left|x-2\right|\ge0\) \(\forall x.\)

\(\Rightarrow-4.\left|x-2\right|\le0\) \(\forall x.\)

\(\Rightarrow10-4.\left|x-2\right|\le10\) \(\forall x.\)

\(\Rightarrow B\le10.\)

Dấu '' = '' xảy ra khi:

\(x-2=0\)

\(\Rightarrow x=0+2\)

\(\Rightarrow x=2.\)

Vậy \(MAX_B=10\) khi \(x=2.\)

Chúc bạn học tốt!

1.\(A=2\left|3x-1\right|-4\)

+Có: \(\left|3x-1\right|\ge0với\forall x\\ \Rightarrow2\left|3x-1\right|-4\ge-4\\ \Leftrightarrow A\ge-4\)

+Dấu ''='' xảy ra khi \(\left|3x-1\right|=0\Leftrightarrow x=\frac{1}{3}\)

+Vậy \(A_{min}=-4\) khi \(x=\frac{1}{3}\)

2.\(B=10-4\left|x-2\right|\)

+Có: \(-4\left|x-2\right|\le0với\forall x\\ \Rightarrow10-4\left|x-2\right|\le10\\ \Leftrightarrow B\le10\)

+Dấu ''='' xảy ra khi \(\left|x-2\right|=0\Leftrightarrow x=2\)

+Vậy \(B_{max}=10\) khi \(x=2\)

1/ Với mọi x ta có :

\(\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|-4\ge-4\)

\(\Leftrightarrow A\ge-4\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{3}\)

Vậy...

2/ Với mọi x ta có :

\(\left|x-2\right|\ge0\)

\(\Leftrightarrow4\left|x-2\right|\ge0\)

\(\Leftrightarrow-4\left|x-2\right|\le0\)

\(\Leftrightarrow10-4\left|x-2\right|\le10\)

\(\Leftrightarrow B\le10\)

Dấu "=" xảy ra \(\Leftrightarrow x=2\)

Vậy..

3/ Với mọi x ta có :

\(\left|x\right|\ge0\)

\(\Leftrightarrow\left|x\right|-3\ge-3\)

\(\Leftrightarrow\frac{6}{\left|x\right|-3}\le-2\)

\(C\le-2\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy...