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a)\(\sqrt{3x-1}\)\(=2\)
⇔\(\text{3x-1=2}^2\)
⇔\(3x=5\)
⇔\(x=\dfrac{5}{3}\)
b)\(\sqrt{x^2-4x+4}\)\(\text{=3x-1}\)
⇔\(\text{x-2=3x-1}\)
⇔\(-2x=1\)
⇔\(x=\)\(\dfrac{-1}{2}\)
c)\(\sqrt{x}\)=2-x sai đề bài
d)\(\sqrt{x^2+4}\)=\(\sqrt{3x+8}\)
⇔\(x^2\)\(\text{+4=3x+8}\)
⇔\(x^2\)\(-3x-4=0\)
⇔\(\left(x+1\right)\left(x-4\right)=0\)
⇔\(\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
\(Q=x^2\left(4-3x\right)=\dfrac{4}{9}.\dfrac{3}{2}x.\dfrac{3}{2}x\left(4-3x\right)\)
\(Q\le\dfrac{1}{27}.\dfrac{4}{9}.\left(\dfrac{3x}{2}+\dfrac{3x}{2}+4-3x\right)^3=\dfrac{256}{243}\)
\(Q_{maxx}=\dfrac{256}{243}\) khi \(\dfrac{3x}{2}=4-3x\Leftrightarrow x=\dfrac{8}{9}\)
Câu 1:
ĐK: \(x\geq -8\)
Đặt \(\sqrt{x+8}=a(a\geq 0)\) thì pt tương đương với:
\((4x+2)a=3x^2+6x+(x+8)=3x^2+6x+a^2\)
\(\Leftrightarrow 3x^2+6x+a^2-4ax-2a=0\)
\(\Leftrightarrow (4x^2-4ax+a^2)-x^2+6x-2a=0\)
\(\Leftrightarrow (2x-a)^2+2(2x-a)-x^2+2x=0\)
\(\Leftrightarrow (2x-a)^2+2(2x-a)+1-(x^2-2x+1)=0\)
\(\Leftrightarrow (2x-a+1)^2-(x-1)^2=0\)
\(\Leftrightarrow (x-a+2)(3x-a)=0\)
\(\bullet \)Nếu \(x-a+2=0\Leftrightarrow x+2=a\Rightarrow (x+2)^2=a^2=x+8\)
\(\Leftrightarrow x^2+3x+4=0\Rightarrow \left[\begin{matrix} x=1\\ x=-4\end{matrix}\right.\) . Ở đây chỉ có TH $x=1$ thỏa mãn còn $x=-4$ bị loại vì $x+2=a\geq 0$
\(\bullet \) Nếu \(3x-a=0\Rightarrow 3x=a\Rightarrow 9x^2=a^2=x+8\)
\(\Leftrightarrow 9x^2-x-8=0\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{-8}{9}\end{matrix}\right.\). Ở đây chỉ có TH $x=1$ thỏa mãn còn $x=-\frac{8}{9}$ loại vì \(9x=a\geq 0\rightarrow x\geq 0\)
Vậy PT có nghiệm duy nhất $x=1$
Câu 2:
ĐK: \(x\geq \frac{-1}{3}\)
Đặt \(\sqrt{3x+1}=a(a\geq 0)\). Khi đó pt đã cho tương đương với:
\(x^2+x+(3x+1)-2x\sqrt{3x+1}=\sqrt{3x+1}\)
\(\Leftrightarrow x^2+x+a^2-2ax=a\)
\(\Leftrightarrow (x^2+a^2-2ax)+(x-a)=0\)
\(\Leftrightarrow (x-a)^2+(x-a)=0\Leftrightarrow (x-a)(x-a+1)=0\)
\(\Rightarrow \left[\begin{matrix} x=a\\ x+1=a\end{matrix}\right.\)
Nếu \(x=a=\sqrt{3x+1}\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2=3x+1\end{matrix}\right.\Rightarrow x=\frac{3+\sqrt{13}}{2}\) (t/m)
Nếu \(x+1=a=\sqrt{3x+1}\Rightarrow \left\{\begin{matrix} x\geq -1\\ (x+1)^2=3x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -1\\ x^2-x=0\end{matrix}\right.\)
\(\Rightarrow x=0\) hoặc $x=1$
Vậy.........
\(1.x^2+2=3x\)
có a+b+c=0(tính Δ cũng cho điểm như vậy)
\(\Rightarrow\left[{}\begin{matrix}x_1=1\\x_2=2\end{matrix}\right.\)
vậy phương trình có nghiệm x=1;x=2
2.\(\left\{{}\begin{matrix}3x=8-2y\\-3y=5-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=8\\4x-3y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x+8y=32\\12x-9y=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
vậy hệ phương trình có 1 nghiệm duy nhất (x;y)=(2;1)
\(1,x^2+2=3x\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4.2=1>0\)
\(\Rightarrow\) Pt có 2 nghiệm pb \(x_1,x_2\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+1}{2}=2\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-1}{2}=1\end{matrix}\right.\)
Vậy \(S=\left\{1;2\right\}\)
\(2,\)\(\left\{{}\begin{matrix}3x=8-2y\\-3y=5-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=8\\4x-3y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x+8y=32\\12x-9y=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=8\\17y=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2.1=8\\y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy hệ pt có 1 nghiệm duy nhất \(\left(x;y\right)=\left(2;1\right)\)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
a) ĐKXĐ: \(-1\leq x\leq 2\)
\(\sqrt{(1+x)(2-x)}=1+2x-2x^2\)
\(\Leftrightarrow \sqrt{2+x-x^2}=1+2x-2x^2=-3+2(2+x-x^2)\)
Đặt \(\sqrt{2+x-x^2}=t(t\geq 0)\). PT trở thành:
\(t=-3+2t^2\)
\(\Leftrightarrow 2t^2-t-3=0\Leftrightarrow (2t-3)(t+1)=0\)
\(\Rightarrow t=\frac{3}{2}\) (do \(t\geq 0)\)
\(\Rightarrow 2+x-x^2=\frac{9}{4}\Rightarrow x^2-x+\frac{1}{4}=0\)
\(\Leftrightarrow (x-\frac{1}{2})^2=0\Rightarrow x=\frac{1}{2}\) (thỏa mãn)
b) ĐK: \(x\geq \frac{1}{3}\)
PT \(\Leftrightarrow \sqrt{(3x-1)+6\sqrt{3x-1}+9}+\sqrt{(3x-1)-6\sqrt{3x-1}+9}=3x+4\)
\(\Leftrightarrow \sqrt{(\sqrt{3x-1}+3)^2}+\sqrt{(\sqrt{3x-1}-3)^2}=3x+4\)
\(\Leftrightarrow \sqrt{3x-1}+3+|\sqrt{3x-1}-3|=3x+4\)
\(\Leftrightarrow |\sqrt{3x-1}-3|=3x-\sqrt{3x-1}+1\)
Nếu \(\sqrt{3x-1}\geq 3\):
\(\Rightarrow \sqrt{3x-1}-3=3x-\sqrt{3x-1}+1\)
\(\Leftrightarrow 3x+4-2\sqrt{3x-1}=0\)
\(\Leftrightarrow (3x-1)-2\sqrt{3x-1}+5=0\)
\(\Leftrightarrow (\sqrt{3x-1}-1)^2+4=0\) (vô lý)
Nếu \(\sqrt{3x-1}< 3\):
\(\Rightarrow 3-\sqrt{3x-1}=3x-\sqrt{3x-1}+1\)
\(\Leftrightarrow 3x=2\Rightarrow x=\frac{2}{3}\) (thỏa mãn)
Vậy...........
a: =>(3x-1)(x-1)<0
=>1/3<x<1
b: =>\(5x^2+17x-5x-17>=0\)
=>(5x+17)(x-1)>=0
=>x>=1 hoặc x<=-17/5
d: =>(x-5)(x-7)<=0
=>5<=x<=7