Giải hộ mình bài này mấy bạn 
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DN
2
8 tháng 11 2021
Bài 4:
\(a,A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ P=A:B=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\\ b,P\sqrt{x}=m-\sqrt{x}+x\\ \Leftrightarrow x-1=m-\sqrt{x}+x\\ \Leftrightarrow m=\sqrt{x}-1\)
AH
Akai Haruma
Giáo viên
23 tháng 7 2021
1.
\(\sqrt{x}-2+x\sqrt{x}-2x=(\sqrt{x}-2)+(x\sqrt{x}-2x)=(\sqrt{x}-2)+x(\sqrt{x}-2)\)
\(=(\sqrt{x}-2)(1+x)\)
2.
\(x-10\sqrt{x}+25=(\sqrt{x})^2-2.5.\sqrt{x}+5^2=(\sqrt{x}-5)^2\)
3.
\(4x+4\sqrt{x}+1=(2\sqrt{x})^2+2.2\sqrt{x}+1=(2\sqrt{x}+1)^2\)
4.
\(9x-6\sqrt{x}+1=(3\sqrt{x})^2-2.3\sqrt{x}+1=(3\sqrt{x}-1)^2\)
AH
Akai Haruma
Giáo viên
23 tháng 7 2021
5.
\(\sqrt{x-1}-5x+5=\sqrt{x-1}-5(x-1)=\sqrt{x-1}(1-5\sqrt{x-1})\)
6.
\(\sqrt{x-3}-2x+6=\sqrt{x-3}-2(x-3)=\sqrt{x-3}(1-2\sqrt{x-3})\)
7.
\(x\sqrt{x}-1=(\sqrt{x})^3-1^3=(\sqrt{x}-1)(x+\sqrt{x}+1)\)
8.
\(x-10\sqrt{x}+21=x-3\sqrt{x}-(7\sqrt{x}-21)\)
\(=\sqrt{x}(\sqrt{x}-3)-7(\sqrt{x}-3)=(\sqrt{x}-7)(\sqrt{x}-3)\)
LO
0
5 tháng 8 2021
22,
1, Đặt √(3-√5) = A
=> √2A=√(6-2√5)
=> √2A=√(5-2√5+1)
=> √2A=|√5 -1|
=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)
=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)
2, Đặt √(7+3√5) = B
=> √2B=√(14+6√5)
=> √2B=√(9+2√45+5)
=> √2B=|3+√5|
=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)
=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)
3,
Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C
=> √2C=√(18+2√17) - √(18-2√17) -\(2\)
=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)
=> √2C=√17+1- √17+1 -\(2\)
=> √2C=0
=> C=0
26,
|3-2x|=2\(\sqrt{5}\)
TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)
3-2x=2\(\sqrt{5}\)
-2x=2\(\sqrt{5}\) -3
x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)
TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)
3-2x=-2\(\sqrt{5}\)
-2x=-2√5 -3
x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)
Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)
6 tháng 8 2021
2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12
3, \(\sqrt{x^2-2x+1}\)=7
⇔ |x-1|=7
TH1: x-1≥0 ⇔ x≥1
x-1=7 ⇔ x=8 (TMĐK)
TH2: x-1<0 ⇔ x<1
x-1=-7 ⇔ x=-6 (TMĐK)
Vậy x=8, -6
4, \(\sqrt{\left(x-1\right)^2}\)=x+3
⇔ |x-1|=x+3
TH1: x-1≥0 ⇔ x≥1
x-1=x+3 ⇔ 0x=4 (KTM)
TH2: x-1<0 ⇔ x<1
x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)
Vậy x=-1
AT
12 tháng 7 2021
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
AT
12 tháng 7 2021
1) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+\sqrt{4.12}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}\right)^2+2.\sqrt{12}+1^2}}=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}=\sqrt{5-\left|\sqrt{4.3}+1\right|}\)
\(=\sqrt{5-\left(2\sqrt{3}+1\right)}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{3}-1}=2\sqrt{2+\sqrt{3}}\)
\(=2\sqrt{\dfrac{4+2\sqrt{3}}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}\)
\(=2.\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
2) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{3}-1\) (như trên)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
NT
các bạn giải hộ mik mấy câu này vs mik đag cần gấp
2
S
22 tháng 8 2021
c, \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
<=> \(C=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
<=> \(C=-5\sqrt{3}:\sqrt{3}=-5\)
e. \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\sqrt{9-5}\)
\(=6+4=10\)
b. \(\left(\sqrt{3}+2\right)^2-\sqrt{75}\)
\(=3+4\sqrt{3}+4-5\sqrt{3}\)
\(=7-\sqrt{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
f. \(\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-2+\sqrt{3}\)
\(=2\sqrt{3}\)
22 tháng 8 2021
c: Ta có: \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\cdot3\sqrt{3}+4\cdot2\sqrt{3}\right):\sqrt{3}\)
\(=2-15+8=-5\)
d: Ta có: \(D=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+3+\sqrt{5}+2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\cdot2=10\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
\(M=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\dfrac{\sqrt{x}-2}{2}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
2. Ta có:
\(\sqrt{x}>0\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>0\) hay \(M>0\)
Lại có: \(M=\dfrac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\dfrac{1}{\sqrt{x}+2}< 1\)
\(\Rightarrow0< M< 1\Rightarrow M>M^2\)
10 tháng 7 2021
1) Ta có: \(M=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{2}\)
\(=\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\)
\(=\dfrac{2\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
3 tháng 7 2021
\(P=\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\) (đk:\(a\ge0;a\ne1\))
\(=\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right).\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)
\(=\dfrac{1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
2) \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)
\(\Leftrightarrow16\sqrt{a}\ge\left(\sqrt{a}+9\right)\left(\sqrt{a}+1\right)\)
\(\Leftrightarrow a-6\sqrt{a}+9\le0\)
\(\Leftrightarrow\left(\sqrt{a}-3\right)^2\le0\)
Dấu "=" xảy ra khi \(\sqrt{a}-3=0\Leftrightarrow a=9\) (tm)
Vậy...
3 tháng 7 2021
1) ĐKXĐ: \(a\ge0;a\ne1\)
\(P=\left[\dfrac{a+\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)\(:\left[\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)
\(\Leftrightarrow P=\left[\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)+2.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right]\)\(:\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow P=\left[\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right].\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)
\(\Leftrightarrow P=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)
\(\Leftrightarrow P=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
2) Có : \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}-\dfrac{\sqrt{a}+9}{8}\ge0\)
\(\Leftrightarrow\dfrac{16\sqrt{a}-\left(\sqrt{a}+9\right).\left(\sqrt{a}+1\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{16\sqrt{a}-a-10\sqrt{a}-9}{8.\left(\sqrt{a}+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{-\left(a-6\sqrt{a}+9\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}\le0\)
Vì \(\sqrt{a}\ge0\Rightarrow8.\left(\sqrt{a}+1\right)>0\) mà \(\left(\sqrt{a}-3\right)^2\) \(\ge0\)
\(\Rightarrow\) \(\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}=0\) \(\Rightarrow\left(\sqrt{a}-3\right)^2=0\) \(\Leftrightarrow\sqrt{a}-3=0\Leftrightarrow\sqrt{a}=3\Leftrightarrow a=9\)
Vậy để\(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\) thì \(a=9\)
\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=6^3-3\left(6-z\right)\left(6-x\right)\left(6-x\right)\)(khai triển tiếp )
và \(0\le x,y,z\le4\Leftrightarrow xyz+\left(4-x\right)\left(4-y\right)\left(4-z\right)\ge0\)(khai triển tiếp )