Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
\(\sqrt{16}=4\)
\(\sqrt{36}=6\)
\(\sqrt{81}=9\)
\(\sqrt{144}=12\)
\(\sqrt{625}=25\)
\(\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}\)
\(\sqrt{\dfrac{36}{25}}=\dfrac{6}{5}\)
\(\sqrt{\dfrac{64}{49}}=\dfrac{8}{7}\)
\(\sqrt{\dfrac{169}{400}}=\dfrac{13}{20}\)
\(\sqrt{11\dfrac{1}{9}}=\sqrt{\dfrac{100}{9}}=\dfrac{10}{3}\)
\(\sqrt{1\dfrac{11}{25}}=\sqrt{\dfrac{36}{25}}=\dfrac{6}{5}\)
\(\sqrt{1\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}=\dfrac{7}{6}\)
Câu 2:
a) \(3.\sqrt{16}-4\sqrt{\dfrac{1}{4}}\)
\(=3.4-4.\dfrac{1}{2}\)
\(=4.\left(3-\dfrac{1}{2}\right)\)
\(=4.\dfrac{5}{2}\)
\(=10\)
b) \(-5\sqrt{\dfrac{9}{16}}+4\sqrt{0,36}-6\sqrt{0,09}\)
\(=-5.\dfrac{3}{4}+4.0,6-6.0,3\)
\(=\dfrac{-15}{4}+\dfrac{12}{5}-\dfrac{9}{5}\)
\(=\dfrac{-75+48-36}{20}=\dfrac{-63}{20}\)
c) \(2.\sqrt{9}-10.\sqrt{\dfrac{1}{25}}\)
\(=2.3-10.\dfrac{1}{5}\)
\(=6-2\)
\(=4\)
d) \(-3\sqrt{\dfrac{25}{16}}+5\sqrt{0,16}-7\sqrt{0,64}\)
\(=-3.\dfrac{5}{4}+5.0,4-7.0,8\)
\(=\dfrac{-15}{4}+2-\dfrac{28}{5}\)
\(=\dfrac{-75+40-28}{20}=\dfrac{-63}{20}\)
e) \(3\sqrt{25}-27\sqrt{\dfrac{4}{81}}\)
\(=3.5-27.\dfrac{2}{9}\)
\(=15-6\)
\(=9\)
f) \(-21\sqrt{\dfrac{100}{49}}+3\sqrt{0,04}-5\sqrt{0,25}\)
\(=-21.\dfrac{10}{7}+3.0,2-5.0,5\)
\(=-30+\dfrac{3}{5}-\dfrac{5}{2}\)
\(=\dfrac{-300+6-25}{10}=\dfrac{-319}{10}\)
h) \(5\sqrt{9}-4\sqrt{\dfrac{1}{16}}+6\sqrt{25}\)
\(=5.3-4.\dfrac{1}{4}+6.5\)
\(=15-1+30\)
\(=14+30\)
\(=44\)
g) \(10\sqrt{\dfrac{9}{25}}-14\sqrt{\dfrac{36}{49}}+24\sqrt{\dfrac{81}{64}}\)
\(=10.\dfrac{3}{5}-14.\dfrac{6}{7}+24.\dfrac{9}{8}\)
\(=6-12+27\)
\(=\left(-6\right)+27=21\)
Câu 3:
a) \(\sqrt{x}=7\)
\(=>x=49\)
b) \(\sqrt{x}=12\)
\(=>x=144\)
c) \(\sqrt{x}=15\)
\(=>x=225\)
d) \(\sqrt{x}=20\)
\(=>x=400\)
e) \(4\sqrt{x}=8\)
\(\sqrt{x}=8:4\)
\(\sqrt{x}=2\)
\(=>x=4\)
f) \(6\sqrt{x}=3\)
\(\sqrt{x}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(=>x=\dfrac{1}{4}\)
g) \(\sqrt{x-1}=1\)
\(x-1=1\)
\(x=1+1\)
\(=>x=2\)
h) \(\sqrt{x+1}=2\)
\(x+1=4\)
\(x=4-1\)
\(=>x=3\)
i) \(\sqrt{x}-2=7\)
\(\sqrt{x}=7+2\)
\(\sqrt{x}=9\)
\(=>x=81\)
j) \(14-\sqrt{x}=12\)
\(\sqrt{x}=14-12\)
\(\sqrt{x}=2\)
\(=>x=4\)
k) \(12-\sqrt{x-1}=2\)
\(\sqrt{x-1}=12-2\)
\(\sqrt{x-1}=10\)
\(x-1=100\)
\(x=100+1\)
\(=>x=101\)
l) \(\sqrt{x+5}+10=20\)
\(\sqrt{x+5}=20-10\)
\(\sqrt{x+5}=10\)
\(x+5=100\)
\(x=100-5\)
\(=>x=95\)
# Wendy Dang
3:
a: ĐKXĐ: x>=0
\(\sqrt{x}=7\)
=>x=7^2=49
b: ĐKXĐ: x>=0
\(\sqrt{x}=12\)
=>x=12^2=144
c: ĐKXĐ: x>=0
\(\sqrt{x}=15\)
=>x=15^2=225
d: ĐKXĐ: x>=0
\(\sqrt{x}=20\)
=>x=20^2=400
e: ĐKXĐ: x>=0
\(4\sqrt{x}=8\)
=>\(\sqrt{x}=2\)
=>x=4
f: ĐKXĐ: x>=0
\(6\cdot\sqrt{x}=3\)
=>\(\sqrt{x}=\dfrac{3}{6}=\dfrac{1}{2}\)
=>x=1/4
g: ĐKXĐ: x>=1
\(\sqrt{x-1}=1\)
=>x-1=1
=>x=2
h: ĐKXĐ: x>=-1
\(\sqrt{x+1}=2\)
=>x+1=4
=>x=3
i: ĐKXĐ: x>=0
\(\sqrt{x}-2=7\)
=>\(\sqrt{x}=9\)
=>x=81
j: ĐKXĐ: x>=0
\(14-\sqrt{x}=12\)
=>\(\sqrt{x}=14-12=2\)
=>x=4
k: ĐKXĐ: x>=1
\(12-\sqrt{x-1}=2\)
=>\(\sqrt{x-1}=10\)
=>x-1=100
=>x=101
i: ĐKXĐ: x>=-5
\(\sqrt{x+5}+10=20\)
=>\(\sqrt{x+5}=10\)
=>x+5=100
=>x=95
4. \(\dfrac{-3}{2}+x-\dfrac{5}{4}=\dfrac{-1}{3}-2x\)
<=> \(\dfrac{-18}{12}+\dfrac{12x}{12}-\dfrac{15}{12}=\dfrac{-4}{12}-\dfrac{24x}{12}\)
<=> -18 + 12x - 15 = -4 - 24x
<=> 12x + 24x = 18 + 15 - 4
<=> 36x = 29
<=> x = \(\dfrac{29}{36}\)
6. \(\dfrac{3}{4}x-\dfrac{3}{2}=\dfrac{5}{6}+\dfrac{3}{8}x\)
<=> \(\dfrac{18x}{24}-\dfrac{36}{24}=\dfrac{20}{24}+\dfrac{9x}{24}\)
<=> 18x - 36 = 20 + 9x
<=> 18x - 9x = 20 + 36
<=> 9x = 56
<=> x = \(\dfrac{56}{9}\)
7. \(3-\left(\dfrac{1}{2}+2x\right)=\dfrac{2}{3}-x\)
<=> \(3-\dfrac{1}{2}-2x=\dfrac{2}{3}-x\)
<=> \(\dfrac{18}{6}-\dfrac{3}{6}-\dfrac{12x}{6}=\dfrac{4}{6}-\dfrac{6x}{6}\)
<=> 18 - 3 - 12x = 4 - 6x
<=> 15 - 4 = 12x - 6x
<=> 11 = 6x
<=> x = \(\dfrac{11}{6}\)
Bài 1:
a) Ta có: \(\widehat{aOd}+\widehat{bOd}=180^o\) (kề bù)
\(\Rightarrow\widehat{aOd}=180^o-\widehat{bOd}=180^o-70^o=110^o\)
Do Oc là phân giác của góc aOd nên:
\(\widehat{cOd}=\dfrac{\widehat{aOd}}{2}=\dfrac{110^o}{2}=55^o\)
b) \(\widehat{cOb}=\widehat{cOd}+\widehat{bOd}\) (kề)
\(\Rightarrow\widehat{cOb}=55^o+70^o=125^o\)
Bài 2:
a) Ta có: \(\widehat{xIz}+\widehat{zIy}=180^o\) (kề bù)
\(\Rightarrow\widehat{xIz}=180^o-40^o=140^o\)
b) Do It là phân giác của góc xIz nên ta có:
\(\widehat{tIz}=\dfrac{\widehat{xIz}}{2}=\dfrac{140^o}{2}=70^o\)
Mà: \(\widehat{tIy}=\widehat{tIz}+\widehat{zIy}\)
\(\Rightarrow\widehat{tIy}=70^o+40^o=110^o\)
Ta có: a//b
=>\(\widehat{A_1}=\widehat{B_3}\)(hai góc so le trong)
mà \(\widehat{A_1}=120^0\)
nên \(\widehat{B_3}=120^0\)
Ta có: \(\widehat{B_3}+\widehat{B_4}=180^0\)(hai góc kề bù)
=>\(\widehat{B_4}+120^0=180^0\)
=>\(\widehat{B_4}=180^0-120^0=60^0\)
Ta có: a//b
=>\(\widehat{D_2}=\widehat{C_4}\)(hai góc so le trong)
mà \(\widehat{D_2}=100^0\)
nên \(\widehat{C_4}=100^0\)
Ta có: \(\widehat{C_2}=\widehat{C_4}\)(hai góc đối đỉnh)
mà \(\widehat{C_4}=100^0\)
nên \(\widehat{C_2}=100^0\)
Bài 5:
Theo hình vẽ, ta có: Bx//ab//Dy
Ta có: Bx//ab
=>\(\widehat{xBC}=\widehat{bCB}\)(hai góc so le trong)
mà \(\widehat{xBC}=40^0\)
nên \(\widehat{bCB}=40^0\)
Ta có: ab//Dy
=>\(\widehat{bCD}=\widehat{yDC}\)(hai góc so le trong)
mà \(\widehat{yDC}=25^0\)
nên \(\widehat{bCD}=25^0\)
\(\widehat{BCD}=\widehat{bCB}+\widehat{bCD}\)
\(=25^0+40^0\)
\(=65^0\)
Bài 6:
Theo hình vẽ, ta có: Ex//ab//Gy
Ta có: Ex//ab
=>\(\widehat{bDE}+\widehat{E}=180^0\)(hai góc trong cùng phía)
=>\(\widehat{bDE}+140^0=180^0\)
=>\(\widehat{bDE}=180^0-140^0=40^0\)
ab//Gy
=>\(\widehat{bDG}=\widehat{yGD}\)(hai góc so le trong)
mà \(\widehat{yGD}=30^0\)
nên \(\widehat{bDG}=30^0\)
\(\widehat{EDG}=\widehat{bDE}+\widehat{bDG}\)
\(=30^0+40^0\)
\(=70^0\)
1,2=6/5
Hiệu số phần bằng nhau là:
6-5=1(phần)
Điểm số của Leicester City là:
11:1*6=66(điểm)
Điểm số của Aston Villa là:
66-11=55(điểm)
Gọi gốc là điểm A, chỗ gãy là B, ngọn đã gãy là điểm C
Xét tam giác ABC vuông tại A có: AB = 6m, BC = 16m - 6m = 10m
=> AB2 + AC2 = BC2 (Định lý Py-ta-go)
Thay: 62 + AC2 = 102
36 + AC2 = 100
AC2 = 100 - 36 = 64
AC = 8 (m)
Vậy khoảng cách từ gốc đến ngọn cây bị gãy là 8 mét
Nếu đúng hãy K cho mình nha
Học tốt nhé
a, Ta có : \(P_{\left(x\right)}=x^2+5x^4-3x^3+x^2+4x^4+3x^2-x+5\)
\(=9x^4-3x^3+5x^2-x+5\)
Ta có : \(Q_{\left(x\right)}=x-5x^3-x^2-x^4+4x^3-x^2+3x-1\)
\(Q_{\left(x\right)}=-x^4-x^3-2x^2+4x-1\)
b, Ta có : \(P_{\left(x\right)}+Q_{\left(x\right)}=9x^4-3x^3+5x^2-x+5-x^4-x^3-2x^2+4x-1\)
\(=8x^4-4x^3+3x^2+3x+4\)
Ta có : \(P_{\left(x\right)}-Q_{\left(x\right)}=9x^4-3x^3+5x^2-x+5+x^4+x^3+2x^2-4x+1\)
\(=10x^4-2x^3+7x^2-5x+6\)
2:
a: \(A=2^4\cdot5-\left[131-\left(13-4\right)^2\right]\)
\(=16\cdot5-\left[131-81\right]\)
=80-50
=30
b: \(B=2^3+3\cdot\left(\dfrac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\dfrac{1}{2}\right]-8\)
\(=8+3-1+\left[4\cdot2\right]-8\)
=8+2
=10
Bài 1:
a: \(A=32,125-\left(6,325+12,125\right)-\left(37+13,675\right)\)
\(=32,125-12,125-6,325-13,675-37\)
=20-20-37
=-37
b: \(B=4,75+\left(-\dfrac{1}{2}\right)^3+0,5^2-3\cdot\dfrac{-3}{8}\)
\(=4,75-0,125+0,25+1,125\)
=5+1
=6
Bài 4:
a) \(2^x=2^5\)
\(\Rightarrow x=5\)
b) \(\left(-7\right)^x=\left(-7\right)^9\)
\(\Rightarrow x=9\)
c) \(4^x=64\)
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3\)
d) \(5^x=625\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\)
3:
a: \(-\dfrac{3}{5}-x=-0,75\)
=>\(x=-\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{3}{20}\)
b: \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
=>\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
=>x=2/5
c: \(-0,15-x=1\dfrac{4}{5}\)
=>\(-0,15-x=1,8\)
=>x=-0,15-1,8=-1,95
d: \(-\dfrac{4}{7}-x=\dfrac{3}{5}\)
=>\(x=-\dfrac{4}{7}-\dfrac{3}{5}\)
=>\(x=\dfrac{-20-21}{35}=-\dfrac{41}{35}\)
a.
\(A\left(x\right)=6x^4-x^3+3x^2-1\)
\(A\left(x\right)\) có bậc 4, hệ số tự do là -1, hệ số cao nhất là 6
\(B\left(x\right)=-2x^4-x^3+2x+1\)
\(B\left(x\right)\) có bậc 4, hệ số tự do là 1, hệ số cao nhất là -2
b.
\(A\left(x\right)+B\left(x\right)=4x^4-2x^3+3x^2+2x\)
\(A\left(x\right)-B\left(x\right)=8x^4+3x^2-2x-2\)
c.
\(S\left(x\right)=4x^4-2x^3+3x^2+2x\)
tại \(x=-1\Rightarrow S\left(-1\right)=4.\left(-1\right)^4-2.\left(-1\right)^3+3.\left(-1\right)^2+2.\left(-1\right)=7\)