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\(a,\Delta=4\left(m-1\right)^2-4\left(-2m-3\right)=4m^2-8m+4+8m+12\\ \Delta=4m^2+16>0\left(đpcm\right)\\ b,\Delta=\left(2m-1\right)^2-4\left(2m-2\right)=4m^2-4m+1-8m+8\\ \Delta=4m^2-12m+9=\left(2m-3\right)^2\ge0\left(đpcm\right)\\ c,Sửa:x^2-2\left(m+1\right)x+2m-2=0\\ \Delta=4\left(m+1\right)^2-4\left(2m-2\right)=4m^2+8m+4-8m+8\\ \Delta=4m^2+12>0\left(đpcm\right)\\ d,\Delta=4\left(m+1\right)^2-4\cdot2m=4m^2+8m+4-8m\\ \Delta=4m^2+4>0\left(đpcm\right)\\ e,\Delta=4m^2-4\left(m+7\right)=4m^2-4m+7=\left(2m-1\right)^2+6>0\left(đpcm\right)\\ f,\Delta=4\left(m-1\right)^2-4\left(-3-m\right)=4m^2-8m+4+12+4m\\ \Delta=4m^2-4m+16=\left(2m-1\right)^2+15>0\left(đpcm\right)\)
Em thử nha,sai thì thôi ạ.
2/ ĐK: \(-2\le x\le2\)
PT \(\Leftrightarrow\sqrt{2x+4}-\sqrt{8-4x}=\frac{6x-4}{\sqrt{x^2+4}}\)
Nhân liên hợp zô: với chú ý rằng \(\sqrt{2x+4}+\sqrt{8-4x}>0\) với mọi x thỏa mãn đk
PT \(\Leftrightarrow\frac{6x-4}{\sqrt{2x+4}+\sqrt{8-4x}}-\frac{6x-4}{\sqrt{x^2+4}}=0\)
\(\Leftrightarrow\left(6x-4\right)\left(\frac{1}{\sqrt{2x+4}+\sqrt{8-4x}}-\frac{1}{\sqrt{x^2+4}}\right)=0\)
Tới đây thì em chịu chỗ xử lí cái ngoặc to rồi..
1.\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)
ĐK \(x\ge-1\)
Nhân liên hợp ta có
\(\left(x+3-x-1\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
<=>\(x^2+\sqrt{\left(x+1\right)\left(x+3\right)}=x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
<=> \(\left(x^2-x\sqrt{x+3}\right)+\left(\sqrt{\left(x+1\right)\left(x+3\right)}-x\sqrt{x+1}\right)=0\)
<=> \(\left(x-\sqrt{x+3}\right)\left(x-\sqrt{x+1}\right)=0\)
<=> \(\orbr{\begin{cases}x=\sqrt{x+3}\\x=\sqrt{x+1}\end{cases}}\)
=> \(x\in\left\{\frac{1+\sqrt{13}}{2};\frac{1+\sqrt{5}}{2}\right\}\)
Vậy \(x\in\left\{\frac{1+\sqrt{13}}{2};\frac{1+\sqrt{5}}{2}\right\}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)\left(x+y-6\right)=0\\y-x-3=0\left(3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\left(y+1\right)\left(1\right)\\x=6-y\left(2\right)\end{matrix}\right.\\y-x-3=0\left(3\right)\end{matrix}\right.\)
\(thế\left(1\right)\left(2\right)vào\left(3\right)\Rightarrow\left(x;y\right)\)
Đặt \(a=x,b=\frac{1}{x}\) thì ta có ab = 1
\(a-b=x-\frac{1}{x}=\frac{x^2-1}{x}=\frac{\left(x-1\right)\left(x+1\right)}{x}\). Vì \(x>1\) nên ta có \(a-b>0\)
\(3\left(a^2-b^2\right)< 2\left(a^3-b^3\right)\)
\(\Leftrightarrow3\left(a-b\right)\left(a+b\right)< 2\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow\left(a^2+ab+b^2\right)>\frac{3}{2}\left(a+b\right)\) (chia cả hai vế cho \(a-b>0\))
\(\Leftrightarrow\left(a^2-\frac{3}{2}a+\frac{9}{16}\right)+\left(b^2-\frac{3}{2}b+\frac{9}{16}\right)+\frac{7}{8}>0\)(vì ab = 1)
\(\Leftrightarrow\left(a-\frac{3}{4}\right)^2+\left(b-\frac{3}{4}\right)^2+\frac{7}{8}>0\) (luôn đúng)
Vậy có đpcm.
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)
Xét \(\Delta=\left[-2\left(m-1\right)\right]^2-4\left(m-3\right)=4\left(m-1\right)^2-4m+12\)
\(=4m^2-8m+4-4m+12\)
\(=4m^2-12m+16\)
\(=\left(2m+3\right)^2+7>0\forall m\)
Do \(\Delta>0\) nên PT trên luôn có hai nghiệm phân biệt \(\forall m\)
(a=1;b=-2(m-1);c=m-3)
Mà pt có 2 nghiệm với mọi m
=> Denta >0
<=>4(m-1)^2-4(m-3)>0
<=>4m^2-8m+4-4m+12>0
<=>4m^2-12m+16>0
<=>4m^2-12m+9+7>0
<=>(2m-3)^2+7>0
Ta thấy: \(\hept{\begin{cases}\left(2m-3\right)^2>0\\7>0\end{cases}\Rightarrow dpcm}\)
=(x^8-x^7+x^6)-(x^6-x^5+x^4)-(x^3-x^2+x)+(x^2-x+1)
=x^6(x^2-x+1)-x^4(x^2-x+1)-x(x^2-x+1)+(X^2-x+1)
=(x^6-x^4-x+1)(x^2-x+1)>0(vì x^6-x^4-x+1>0 và x^2-x+1>0)