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a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4
=>2*căn(x+5)=4
=>căn (x+5)=2
=>x+5=4
=>x=-1
b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
=>2*căn x-1=16
=>x-1=64
=>x=65
c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)
TH1: \(x\ge3\)
\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)
TH2: \(2\le x< 3\)
\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)
TH3: \(0\le x< 2\)
\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
TH4: \(x< 0\)
\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)
`a)sqrt{x^2-2x+1}=2`
`<=>sqrt{(x-1)^2}=2`
`<=>|x-1|=2`
`**x-1=2<=>x=3`
`**x-1=-1<=>x=-1`.
Vậy `S={3,-1}`
`b)sqrt{x^2-1}=x`
Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)
`<=>x>=1`
`pt<=>x^2-1=x^2`
`<=>-1=0` vô lý
Vậy pt vô nghiệm
`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`
`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`
`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4`
`<=>x=9(tmđk)`
Vậy `S={9}.`
`d)x-5sqrt{x-2}=-2(x>=2)`
`<=>x-2-5sqrt{x-2}+4=0`
Đặt `a=sqrt{x-2}`
`pt<=>a^2-5a+4=0`
`<=>a_1=1,a_2=4`
`<=>sqrt{x-2}=1,sqrt{x-2}=4`
`<=>x_1=3,x_2=18`,
`e)2x-3sqrt{2x-1}-5=0`
`<=>2x-1-3sqrt{2x-1}-4=0`
Đặt `a=sqrt{2x-1}(a>=0)`
`pt<=>a^2-3a-4=0`
`a-b+c=0`
`<=>a_1=-1(l),a_2=4(tm)`
`<=>sqrt{2x-1}=4`
`<=>2x-1=16`
`<=>x=17/2(tm)`
Vậy `S={17/2}`
d.
ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:
$a^2+2-5a=-2$
$\Leftrightarrow a^2-5a+4=0$
$\Leftrightarrow (a-1)(a-4)=0$
$\Rightarrow a=1$ hoặc $a=4$
$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$
$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)
e. ĐKXĐ: $x\geq \frac{1}{2}$
Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:
$a^2+1-3a-5=0$
$\Leftrightarrow a^2-3a-4=0$
$\Leftrightarrow (a+1)(a-4)=0$
Vì $a\geq 0$ nên $a=4$
$\Leftrightarrow \sqrt{2x-1}=4$
$\Leftrightarrow x=\frac{17}{2}$
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
Bài 1:
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
Bài 2:
a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)
b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)
Lời giải:
a)
Ta có: \(A=4x^2-x-2=(2x)^2-2.2x.\frac{1}{4}x+(\frac{1}{4})^2-\frac{33}{16}\)
\(=(2x-\frac{1}{4})^2-\frac{33}{16}\)
Vì \((2x-\frac{1}{4})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A\ge 0-\frac{33}{16}=-\frac{33}{16}\)
Vậy GTNN của $A$ là $\frac{-33}{16}$ khi $x=\frac{1}{8}$
b)
\(B=\frac{2x^2+6x-3}{5}=\frac{2(x^2+3x+\frac{9}{4})-\frac{15}{2}}{5}\)
\(=\frac{2(x+\frac{3}{2})^2-\frac{15}{2}}{5}\geq \frac{2.0-\frac{15}{2}}{5}=\frac{-3}{2}\)
Vậy \(B_{\min}=\frac{-3}{2}\Leftrightarrow (x+\frac{3}{2})^2=0\Leftrightarrow x=\frac{-3}{2}\)
c)
\(C=x^4+4x-1\)
\(=x^4-2x^2+1+2x^2+4x-2\)
\(=(x^2-1)^2+2(x^2+2x+1)-4\)
\(=(x^2-1)^2+2(x+1)^2-4\)
\(=(x-1)^2(x+1)^2+2(x+1)^2-4=(x+1)^2[(x-1)^2+2]-4\)
Thấy rằng:
\((x+1)^2\geq 0; (x-1)^2+2>0\Rightarrow (x+1)^2[(x-1)^2+2]\geq 0\)
\(\Rightarrow C\geq 0-4=-4\)
Vậy $C_{\min}=-4$ khi \((x+1)^2=0\Leftrightarrow x=-1\)
d)
\(D=4x^2+\frac{9}{x^2}=(2x)^2+(\frac{3}{x})^2-2.2x.\frac{3}{x}+12\)
\(=(2x-\frac{3}{x})^2+12\geq 0+12=12\)
Vậy $D_{\min}=12$ khi \(2x-\frac{3}{x}=0\Leftrightarrow x=\pm \sqrt{\frac{3}{2}}\)