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b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{2}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=-\dfrac{1}{10}\)
\(\Rightarrow x=\dfrac{1}{4}:\left(-\dfrac{1}{10}\right)\)
\(\Rightarrow x=-\dfrac{3}{2}\)
\(a)18^{20}.45^5.5^{25}.8^{10}\)
\(=\left(2.3^2\right)^{20}.\left(3^2.5\right)^5.5^{25}.\left(2^3\right)^{10}\)
\(=2^{20}.3^{40}.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=7776^{10}.125^{10}\)
\(=972000^{10}\)
\(b)\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(xy\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^6.x^3\)
\(=x^{33}.y^{25}\)
\(=x^{25}.y^{25}.x^8\left(?\right)\)
\(c)2^7.3^8.4^9.9^8\)
\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)\(\left(?\right)\)
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
Bài 1:
a) Đề ko rõ, coi lại
b) \(75^{20}=45^{10}.5^{30}\)
\(\Leftrightarrow\left(75^2\right)^{10}=45^{10}.\left(5^3\right)^{10}\)
\(\Leftrightarrow5625^{10}=45^{10}.125^{10}\)
\(\Leftrightarrow5625^{10}=\left(45.125\right)^{10}\)
\(\Leftrightarrow5625^{10}=5625^{10}\)
\(\Rightarrow75^{20}=45^{10}.5^{30}\left(đpcm\right)\)
Bài 2:
a) \(\frac{x}{-4}=\frac{-3}{5}\)
\(\Rightarrow x.5=-4.\left(-3\right)\)
\(\Rightarrow x.5=12\)
\(\Rightarrow x=\frac{12}{5}=2,4\)
b) c) d) Làm tương tự câu a. Bn tự lm cho nhớ
e) \(30.5x=4.12\)
\(\Rightarrow150x=48\)
\(\Rightarrow x=\frac{48}{150}=0,32\)
f) g) Làm tương tự câu e. Bn tự lm cho nhớ
a) 106 - 57
= 26 . 56 - 57
= 56 . (26 - 5)
= 56 . (64 - 5)
= 56 . 59 chia hết cho 59
=> đpcm
b) 817 - 279 - 913
= (34)7 - (33)9 - (32)13
= 328 - 327 - 326
= 326 .(32 - 3 - 1)
= 326 . (9 - 3 - 1)
= 324 . 32 . 5
= 324 . 9 . 5
= 324 . 45 chia hết cho 45
=> đpcm
c) 87 - 218
= (23)7 - 218
= 221 - 218
= 218 . (23 - 1)
= 218 (8 - 1)
= 217 . 2 . 7
= 217 . 14 chia hết cho 14
=> đpcm
d) 109 + 108 + 107
= 107 . (102 + 10 + 1)
= 57 . 27 . (100 + 10 + 1)
= 57 . 26 . 2 . 111
= 57 . 26 . 222 chia hết cho 222
=> đpcm
Ta có:\(8^7-2^{18}=\left(2^3\right)^7-2^{18}\)\(=2^{21}-2^{18}=2^{17}\cdot2^4-2^{17}\cdot2=2^{17}\cdot\left(2^4-1\right)=2^{17}\cdot14\)\(⋮14\)
\(\Rightarrow8^7-2^{18}⋮14\)
(ĐPCM)
1, Ta có: \(8^7-2^{18}=2^{21}-2^{18}=2^{17}\left(2^4-2\right)=2^{17}.14⋮14\)
2, Đặt: \(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=k\)
\(\Rightarrow a=2k;b=5k;c=7k\)
Nên: \(A=\dfrac{a-b+c}{a+2b-c}=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{3k}{5k}=\dfrac{3}{5}\)
Vậy...
1.
a) \(\frac{3}{7}+\frac{-5}{2}+\frac{-3}{5}\\ =\frac{30}{70}+\frac{-175}{70}+\frac{-42}{70}\\ =\frac{30-175-42}{70}\\ =\frac{-187}{70}\)
b) \(\frac{-8}{18}-\frac{15}{27}\\ =\frac{-4}{9}-\frac{5}{9}\\ =\frac{-9}{9}=-1\)
c) \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\\ =\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\\ =\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\\ =\frac{56+20-49}{70}\\ =\frac{27}{70}\)
2.
a) \(x+\frac{1}{4}=\frac{4}{3}\\ x=\frac{4}{3}-\frac{1}{4}\\ x=\frac{16}{12}-\frac{3}{12}\\ x=\frac{13}{12}\)
Vậy \(x=\frac{13}{12}\)
b) \(-x-\frac{2}{3}=\frac{-6}{7}\\ -x=\frac{-6}{7}+\frac{2}{3}\\ -x=\frac{-18}{21}+\frac{14}{21}\\ -x=-\frac{4}{21}\\ x=\frac{4}{21}\)
Vậy \(x=\frac{4}{21}\)
c) \(x^2=16\\ x^2=4^2=\left(-4\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{4;-4\right\}\)
d) \(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\\ \frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\\ \frac{5}{3}x=\frac{15}{21}-\frac{14}{21}\\ \frac{5}{3}x=\frac{1}{21}\\ x=\frac{1}{21}:\frac{5}{3}\\ x=\frac{1}{21}\cdot\frac{3}{5}\\ x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}\)
3.
a) Xét △AKB và △AKC có:
AB = AC
KB = KC
AK: cạnh chung
\(\Rightarrow\text{△AKB = △AKC (c.c.c) }\)
b) \(\text{△AKB = △AKC }\)
\(\Rightarrow\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)
Mà \(\widehat{AKB}+\widehat{AKC}=180^o\) (2 góc kề bù)
\(\Rightarrow\widehat{AKB}=\widehat{AKC}=90^o\\ \Rightarrow AK\perp BC\)
Câu 3:
a/ Xét ΔAKB và ΔAKC có:
AB = AC (GT)
\(\widehat{BAK}=\widehat{CAK}\left(GT\right)\)
AK: cạnh chung
=> ΔAKB = ΔAKC (c.g.c)
b/ VìΔAKB = ΔAKC (câu a)
\(\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)
Mà 2 góc này lại là hai góc kề bù
=> \(\widehat{AKB}=\widehat{AKC}=180^0:2=90^0\)
=> AK ⊥BC
Cau 2:
a) \(x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
b) \(-x-\frac{2}{3}=-\frac{6}{7}\)
=> \(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\)
=> \(x=\frac{4}{21}\)
c) x2 = 16
=> x = 4 hoặc x =-4
a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)