Giúp e vs a
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H9
HT.Phong (9A5)
CTVHS
19 tháng 8 2023
a) \(A=\left(2\sqrt{12}-\sqrt{75}+\dfrac{1}{2}\sqrt{48}\right):\sqrt{3}\)
\(A=\left(4\sqrt{3}-5\sqrt{3}+2\sqrt{3}\right):\sqrt{3}\)
\(A=\sqrt{3}:\sqrt{3}\)
\(A=1\)
b) \(B=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(B=\left|2-\sqrt{5}\right|-\left|\sqrt{5}+1\right|\)
\(B=-2+\sqrt{5}-\sqrt{5}-1\)
\(B=-3\)
c) \(C=\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{3+\sqrt{7}}\)
\(C=\dfrac{3\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{4\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(C=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\dfrac{4\left(3-\sqrt{7}\right)}{2}\)
\(C=\sqrt{7}+2-2\left(3-\sqrt{7}\right)\)
\(C=\sqrt{7}+2-6+2\sqrt{7}\)
\(C=3\sqrt{7}-4\)
d) \(D=3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\)
\(D=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
\(D=5\sqrt{2a}-3a\sqrt{2a}-2\sqrt{2a}\)
\(D=3\sqrt{2a}-3a\sqrt{2a}\)
e) \(E=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)
\(E=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(E=\left(\sqrt{3}+1\right)-\dfrac{2\left(\sqrt{3}+1\right)}{2}\)
\(E=\left(\sqrt{3}+1\right)-\left(\sqrt{3}+1\right)\)
\(E=0\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2023
Lời giải:
a.
\(A=2\sqrt{\frac{12}{3}}-\sqrt{\frac{75}{3}}+\frac{1}{2}\sqrt{\frac{48}{3}}=2\sqrt{4}-\sqrt{25}+\frac{1}{2}\sqrt{16}\)
\(2.2-5+\frac{1}{2}.4=1\)
b.
\(B=|2-\sqrt{5}|-|\sqrt{5}+1|=\sqrt{5}-2-(\sqrt{5}+1)=-3\)
c.
\(C=\frac{3(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}-\frac{4(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}\)
\(=\frac{3(\sqrt{7}+2)}{7-2^2}-\frac{4(3-\sqrt{7})}{3^2-7}\)
\(=\frac{3(\sqrt{7}+2)}{3}-\frac{4(3-\sqrt{7})}{2}=\sqrt{7}+2-2(3-\sqrt{7})=-4+3\sqrt{7}\)
e.
\(E=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}-\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\sqrt{3}+1-\frac{2(\sqrt{3}+1)}{3-1^2}=(\sqrt{3}+1)-(\sqrt{3}+1)=0\)
giúp e vs
22 tháng 10 2021
1: \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
2: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
3: \(\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
19 tháng 6 2023
a: CH=16^2/24=256/24=32/3(cm)
BC=24+32/3=104/3cm
AC=căn 32/3*104/3=16/3*căn 13(cm)
b: BC=12^2/6=144/6=24cm
CH=24-6=18cm
AC=căn 18*24=12*căn 3(cm)
MA
2
23 tháng 12 2021
g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
23 tháng 12 2021
\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
TN
1
DT
Đỗ Thanh Hải
CTVVIP
11 tháng 8 2021
1 I wish my students studied hard
2 I wish I could come to the party
3 I wish my grandparents didn't live too far from me
4 Nga wished she were in HN now
5 This room is cleaned everyday
LK
3
18 tháng 2 2023
| Số | Phần mười | Phần trăm | Phần nghìn | Phần chục nghìn |
| 0,0236891 | 0,0 | 0,02 | 0,024 | 0,0237 |
| 2,1738999 | 2,2 | 2,17 | 2,174 | 2,1739 |
H9
HT.Phong (9A5)
CTVHS
18 tháng 2 2023
| Số | Tròn phần mười | Tròn phần trăm | Tròn phần nghìn | Tròn phần vạn |
| \(0,0236891\) | \(0,0\) | \(0,02\) | \(0,024\) | \(0,0237\) |
| \(2,1738999\) | \(2,2\) | \(2,17\) | \(2,174\) | \(2,1739\) |
18 tháng 7 2021
a) Ta có: \(x-\dfrac{1}{2}=\left|\dfrac{3}{7}\right|\)
nên \(x-\dfrac{1}{2}=\dfrac{3}{7}\)
hay \(x=\dfrac{3}{7}+\dfrac{1}{2}=\dfrac{6}{14}+\dfrac{7}{14}=\dfrac{13}{14}\)
b) Ta có: |x-1|=0
nên x-1=0
hay x=1
c) Ta có: \(\left|x+1\right|\ge0\forall x\)
\(\left|y-2\right|\ge0\forall y\)
Do đó: \(\left|x+1\right|+\left|y-2\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi x=-1 và y=2
18 tháng 7 2021
d) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}\)
mà x-y=-4
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{-4}{-2}=2\)
Do đó: x=6; y=10
e) Ta có: 3x=4y
nên \(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}\)
Đặt \(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}k\\y=\dfrac{1}{4}k\end{matrix}\right.\)
Ta có: xy=48
nên \(\dfrac{1}{3}k\cdot\dfrac{1}{4}k=48\)
\(\Leftrightarrow k^2\cdot\dfrac{1}{12}=48\)
\(\Leftrightarrow k^2=48\cdot12=576\)
hay \(k\in\left\{24;-24\right\}\)
Trường hợp 1: k=24
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}k=\dfrac{1}{3}\cdot24=8\\y=\dfrac{1}{4}k=\dfrac{1}{4}\cdot24=6\end{matrix}\right.\)
Trường hợp 2: k=-24
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}k=\dfrac{1}{3}\cdot\left(-24\right)=-8\\y=\dfrac{1}{4}k=\dfrac{1}{4}\cdot\left(-24\right)=-6\end{matrix}\right.\)
\(D=8+8^3+8^5+...+8^{2x+1}\)
\(\Rightarrow64D=8^3+8^5+...+8^{2x+3}\)
\(\Rightarrow63D=64D-D=8^3+8^5+...+8^{2x+3}-8-8^3-...-8^{2x+1}=8^{2x+3}-8\)
\(\Rightarrow8^{2x+3}-8+8=8^{51}\)
\(\Rightarrow8^{2x+3}=8^{51}\Rightarrow2x+3=51\Rightarrow2x=48\Rightarrow x=24\)