\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

nhỏ quá bn !

b) Thay y=0 vào (d1), ta được:

\(\dfrac{1}{2}x+2=0\)

\(\Leftrightarrow x\cdot\dfrac{1}{2}=-2\)

hay x=-4

Vậy: A(-4;0)

Thay y=0 vào (d2), ta được:

\(2-x=0\)

hay x=2

Vậy: B(2;0)

Phương trình hoành độ giao điểm của (d1) và (d2) là:

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

\(\Leftrightarrow x=0\)

Thay x=0 vào (d2), ta được:

\(y=-0+2=2\)

Vậy: C(0;2)

\(AB=\sqrt{\left(-4-2\right)^2+\left(0-0\right)^2}=6\)

\(AC=\sqrt{\left(-4-0\right)^2+\left(0-2\right)^2}=2\sqrt{5}\)

\(BC=\sqrt{\left(2-0\right)^2+\left(0-2\right)^2}=2\sqrt{2}\)

c) Chu vi tam giác ABC là:

\(C_{ABC}=AB+AC+BC=6+2\sqrt{5}+2\sqrt{2}\left(cm\right)\)

\(\Leftrightarrow P_{ABC}=\dfrac{C_{ABC}}{2}=3+\sqrt{5}+\sqrt{2}\)

Diện tích tam giác ABC là:

\(S_{ABC}=\sqrt{P\cdot\left(P-AB\right)\left(P-AC\right)\left(P-BC\right)}\)

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

\(=6\left(cm^2\right)\)

a:Thay x=1 và y=-1 vào (d), ta được:

2a-4=-1

hay \(a=\dfrac{3}{2}\)

a) Ta có: \(A\left(1;-1\right)\in\left(d\right)\Rightarrow y_A=\left(2a-1\right)x_A-3\)

\(\Rightarrow-1=\left(2a-1\right).1-3\Rightarrow2a-1=2\)

Vậy \(\left(d\right):y=2x-3\)

b) Ta có: \(\left(d'\right)\perp\left(d\right)\Leftrightarrow a.a'=-1\)

\(\Leftrightarrow a'.2=-1\Leftrightarrow a'=-\dfrac{1}{2}\Leftrightarrow\left(d'\right):y=-\dfrac{1}{2}x+b\)

Ta có: \(\left(d'\right)\) cắt trục tung tại điểm B có tung độ là \(\dfrac{4}{3}\)

\(\Leftrightarrow b=\dfrac{4}{3}\)

Vậy \(\left(d'\right):y=-\dfrac{1}{2}x+\dfrac{4}{3}\)

a, Vì (d) đi qua A(1;-1) nên \(x=1;y=-1\)

\(\Leftrightarrow2a-1-3=-1\\ \Leftrightarrow a=\dfrac{3}{2}\)

Vậy \(\left(d\right):y=\left(2\cdot\dfrac{3}{2}-1\right)x-3=2x-3\)

\(b,\) Gọi \(\left(d'\right):y=cx+d\left(c\ne0\right)\) là đt cần tìm

\(\left(d'\right)\perp\left(d\right)\Leftrightarrow2c=-1\Leftrightarrow c=-\dfrac{1}{2}\)

\(\left(d'\right)\) cắt trục tung tại điểm B có tung độ \(\dfrac{4}{3}\Leftrightarrow x=0;y=\dfrac{4}{3}\Leftrightarrow d=\dfrac{4}{3}\)

Vậy \(\left(d'\right):y=-\dfrac{1}{2}x+\dfrac{4}{3}\)

c, PTHDGD của \(\left(d\right)\) và \(\left(d'\right)\) là \(-\dfrac{1}{2}x+\dfrac{4}{3}=2x-3\Leftrightarrow\dfrac{5}{2}x=\dfrac{13}{3}\Leftrightarrow x=\dfrac{26}{15}\\ \Leftrightarrow y=\dfrac{7}{15}\Leftrightarrow C\left(\dfrac{26}{15};\dfrac{7}{15}\right)\)

d, Ta có \(B\left(0;\dfrac{4}{3}\right)\Leftrightarrow BC=\sqrt{\left(\dfrac{26}{15}-0\right)^2+\left(\dfrac{7}{15}-\dfrac{4}{3}\right)^2}=\dfrac{13\sqrt{5}}{15}\)

\(AC=\sqrt{\left(\dfrac{26}{15}-1\right)^2+\left(\dfrac{7}{15}+1\right)^2}=\dfrac{11\sqrt{5}}{15}\)

Vì \(\Delta ABC\) vuông tại C nên \(S_{ABC}=\dfrac{1}{2}AC\cdot BC=\dfrac{1}{2}\cdot\dfrac{13\sqrt{5}}{15}\cdot\dfrac{11\sqrt{5}}{15}=\dfrac{143}{90}\left(đvdt\right)\)

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

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

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

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

b:

\(\dfrac{x}{12}=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

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

\(\Leftrightarrow\dfrac{x}{12}=\dfrac{1}{3}\)

=>x=36

Khi x=36 thì \(A=36-6+1=37-6=31\)

c: \(B=\dfrac{2\sqrt{x}}{A}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)

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

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

\(=\dfrac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

=>B<2

\(2\sqrt{x}>0;x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

=>B>0

=>0<B<2

a/ Pt có 2 nghiệm phân biệt

\(\to\Delta'=(-m)^2-1.(-8m-16)=m^2+8m+16=(m+4)^2>0\\\to m+4>0\quad or\quad m+4<0\\\to m>-4\quad or\quad m<-4\)

b/ Theo Vi-ét:

\(\begin{cases}x_1+x_2=2m\\x_1x_2=-8m-16\end{cases}\)

\(x_1^2+x_2^2=5\\\leftrightarrow x_1^2+2x_1x_2+x_2^2-2x_1x_2=5\\\leftrightarrow (x_1+x_2)^2-2x_1x_2=5\\\leftrightarrow (2m)^2-2.(-8m-16)=5\\\leftrightarrow 4m^2+16m+32=5\\\leftrightarrow 4(m^2+4m+8)=5\\\leftrightarrow 4(m+2)^2+16=5\\\leftrightarrow 4(m+2)^2+11=0(\text{vô lý})\\\to m\in\varnothing\)

Vậy không có giá trị m thỏa mãn

1a. 

\(2x^2+7xy+5y^2-5y-2x\)

\(=2x^2+5xy+2xy+5y^2-5y-2x\)

\(=x\left(2x+5y\right)+y\left(2x+5y\right)-\left(2x+5y\right)\)

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