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giúp mình với nha mọi ngườiBài 1 : Đồ thị đi qua điểm M(4;-3) \(\Rightarrow\) y=-3 x=4. Ta được:
\(-3=4a+b\)
Đồ thị song song với đường d \(\Rightarrow\) \(a=a'=-\dfrac{2}{3}\) Ta được:
\(-3=4.-\dfrac{2}{3}+b\) \(\Rightarrow\) \(b=-\dfrac{1}{3}\)
Vậy: \(a=-\dfrac{2}{3};b=-\dfrac{1}{3}\)
b) (P) đi qua 3 điểm A B O, thay tất cả vào (P), ta được hpt:
\(\hept{\begin{cases}a+b+c=1\\a-b-c=-3\\0+0+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=-1\\b=2\\c=0\end{cases}}}\)
Bài 2 : Mình ko biết vẽ trên này, bạn theo hướng dẫn rồi tự làm nhé
Đồ thị có \(a< 0\) \(\Rightarrow\) Hàm số nghịch biến trên R
\(\Rightarrow\) Đồ thị có đỉnh \(I\left(1;4\right)\)
Chọn các điểm:
x 1 3 -1 2 -2
y 4 0 0 3 -5
Câu a:
ĐKXĐ: \(x\neq \pm 3\)
\(\left|\frac{x+5}{-x^2+9}\right|=2\Rightarrow \left[\begin{matrix} \frac{x+5}{-x^2+9}=2\\ \frac{x+5}{-x^2+9}=-2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x+5=2(-x^2+9)\\ x+5=-2(-x^2+9)\end{matrix}\right.\Rightarrow \left[\begin{matrix} 2x^2+x-13=0\\ 2x^2-x-23=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1\pm \sqrt{105}}{4}\\ x=\frac{1\pm \sqrt{185}}{4}\end{matrix}\right.\) (đều thỏa mãn )
Vậy.......
Câu b:
ĐKXĐ: \(x< 2\)
Ta có: \(\frac{4}{\sqrt{2-x}}-\sqrt{2-x}=2\)
\(\Rightarrow 4-(2-x)=2\sqrt{2-x}\)
\(\Leftrightarrow 4=(2-x)+2\sqrt{2-x}\)
\(\Leftrightarrow 5=(2-x)+2\sqrt{2-x}+1=(\sqrt{2-x}+1)^2\)
\(\Rightarrow \sqrt{2-x}+1=\sqrt{5}\) (do \(\sqrt{2-x}+1>0\) )
\(\Rightarrow \sqrt{2-x}=\sqrt{5}-1\)
\(\Rightarrow 2-x=6-2\sqrt{5}\)
\(\Rightarrow x=-4+2\sqrt{5}\) (thỏa mãn)
Vậy...........
a: =>4x+12<=2x-1
=>2x<=-13
=>x<=-13/2
b: =>x^2-2x+1+4<0
=>(x-1)^2+4<0(loại)
c: =>(x-2+x+3)/(x+3)<0
=>(2x+1)/(x+3)<0
=>-3<x<-1/2
\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)
\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)
\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)
\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)
a) \(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)
Đặt \(\sqrt{x^2-3x+3}=a;\sqrt{x^2-3x+6}=b\left(a;b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\b^2-a^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\\left(b+a\right)\left(b-a\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+a=3\\b-a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\) (nhận)
\(\Rightarrow\sqrt{x^2-3x+3}=1\)
\(\Leftrightarrow x^2-3x+3=1\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\) (nhận)
b) \(\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)
Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\\left(b^2+2b+1\right)+b^2-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\2\left(b-1\right)\left(b+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) (vì \(b+2>0\)) (nhận)
\(\Rightarrow\sqrt{2+x-x^2}=1\)
\(\Leftrightarrow2+x-x^2=1\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\) (nhận)
d) \(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}=2x+\dfrac{1}{2x}+4\)
\(\Leftrightarrow2\left(x+\dfrac{1}{4x}\right)+4=5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)\)
\(\Leftrightarrow2\left[\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-1\right]-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+4=0\)
\(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+2=0\)
Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)
\(\Rightarrow2a^2-5a+2=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(\text{nhận}\right)\\a=\dfrac{1}{2}\left(\text{loại}\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{2+\sqrt{2}}{2}\\\sqrt{x}=\dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\) (nhận)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\) (nhận)
b)\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\\sqrt{x^2-3x+8}=x-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-3x+8=\left(x-4\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-3x+8=x^2-8x+16\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\5x=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x=\dfrac{8}{5}\left(loại\right)\end{matrix}\right.\)=> pt vô nghiệm
c)\(\left\{{}\begin{matrix}8-x\ge0\\x^2-5x-2=\left(8-x\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x^2-5x-2=x^2-16x+64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\11x=66\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=6\left(nhận\right)\end{matrix}\right.\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
a)x=-2;0;2
b)x=-1,5;-0,5;0;1;1,5;2,5
c)x=-9;-4;0;2;6;11
d)x=-8;2;4;14