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a) \(1:\overline{0,abc}=a+b+c\)
\(\Rightarrow\dfrac{1}{\overline{abc}}=\dfrac{a+b+c}{1000}\)
\(\Rightarrow\overline{abc}\left(a+b+c\right)=1000\)
Mà 0 < a + b + c < 28 nên a + b + c \(\in\) {1; 2; 4; 5; 8; 10; 20; 25}. Mà \(\overline{abc}\ge100\) nên a + b + c \(\le\) 10, do đó a + b + c \(\in\) {1; 2; 4; 5; 8; 10}. Thử từng trường hợp ta được đáp án đúng là a + b + c = 8 và \(\overline{abc}\) = 125
1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
1) \(\left(9^9\right)^{2013}\)= 10269363159364666440076552322773341581561034085240554413684171629845226550910869063141084455165024846467308031862801839537350602585807388907790165677831287422774432660306450530003706882130019126660033621304145739244276173577048090504994820917529469442173652905242934472777858750567472632994664600381934224746675284242716804187707473971153049296389564538282393321100521850729158342672916978486633073346395087524709304026115423816203365757494638423131935882476286148041225377521573071731453557120367321995775004742604569764745022389412766013722532450077367619939069300519001702898185102392773927389960480888542356324726363237536898205586978830302184325193226223435916070961038034935786871565694168032483034776261863802471075705726878653433383001001189241926035182758070542398573188268383074169109020402590360496218759242201271963792394715618265594345634230758007244699004003000401590521959773595723533039737036430015710879179131370760647094133072554170794993632842471406497462695365166916803272574522454401382663974485565680530010978750425197889269057395033275863668478654934441334494555064318484689342316306971521024595876939555467943409513599739742465719710957307401039466505018857934554613930415045936664298639272058657312601916520149572941057253546060280658091085857108287350235860520376248626158812551702239866122771402598673086936929135243309297996461647086887656015121093133495745098227813854645587494331845951709269358587499740880686161437051001446721645931603701931366046756571915591346082194099535179864942435147889719664866893951993209328180552969033445416386172074158156509068184846110009877655498411796133585929465285105476632644661698885141470189436283199349798153583068536942505793691702852246620602269418445330834508954131444268765759312479343419904740139320879242064290138393396190814854006875023217633358501559386869629903562803482598907058580834642187008732774069291138122707731009317247214463199502007349382592744206845610622073119291353793177956259701743316926165329688122906721927196323010889181055169806499566546884164914042278508330036064549558133226697031247070510887763306579421433675607558954912396327853467424003335216349883637063258300867587330031070320552690888583962060709425761455244473416175295550790206629899652326841562128125494362697380378913996157037213809010909152617053065047965873644302701915161491422477028822914991812751244014648364815652852259663562101505343929698300364745277267393347355428142967482152321747112276920645950373078036691708170463157769001081433039723940115958277368318945023698370418990114114623681030598771547893253242183396733689941466450154464716467140441700170890131070394317235669249736167939422225531914712053400394591025170046527933941931808727717700810490226657457458014925192262802223793377911267650955266657089009585212112836905894381397018270698103556284576894624491741924724548232777077039317695115234021720883233465113399660643038825392305224594945823567653088326327442095353311628349624602121813895038502370886964075117719039885809760171422727129924473839457315768243597403319870636550055160900303769922718752206531201831705424385675834623470898120798414884603237556758496483422249797988913495971144948857810070808962140027449957839158509072309335228612816014153586809180977765327121627937134049967684345369108329599698221680897904237253646696104638289317058937956786704502654705018578331925049052381574371364079244827076900746007044670044607514934428774185406569688113571812978834960339563464520445275203854387799426090303262175550913985879685323013395273140584906121284898600419987993686188204435391094252218471390818917130390872182868519308994839897218982949442429019573247952912905380490755419913598457819276169417786284482347581370093174347981877489100149059409603635202204843390807300768332120719828797936653584404544694348383212549192087418173867785361221768506688864308755986946608953282003111974359205430482715512293489410742551889057944409965962731729135907369164794520884407474498460942159861999051690799986820439014933471232036918567390365835132305185662258913590669721711271035876498544692676850173083777815138713451735852959497582505542139720996338872994246201493700854225531805769779199297405335600737006903257207290820931044945024227595231128387120276064384227546402934361068266072587525725727012002788329077620146531366428926553058456985976818503072684025934586637898483958234508662818031180715524520776171094013494021013676728110150423914944710134238003487063081238423668330925015539056597900840885380931769197169725833541445689013104266424340197869967258623982371657927554051872347209361532830788078019771804179098819400418948649540270834597079029891053990824778600110747558315677420029212621805618132160031130257415664172691492945292697559304231368145501988941653172710920650443181254274948908249495935867675652007874393961066550920282780133604505587836446569409476792952876000047659924818891904298270222076421357886611744774356481805662861913303332953231470607411006298630956870297224099368538952834326914631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1)
a) Theo đề ta có:
(99)2013 = 99.2013= 918117
Ta có:
91 = 9 có chữ số tận cùng là 9
92 = 81 có chữ số tận cùng là 1
và 918117 = 92.9058. 9 = (92)9058 .9 ( mình giải thích thêm là mình nhân 9 để cùng cơ số và cộng các tử lại là 2.9058 + 1 = 18117)
Vì 92 có chữ số tận cùng là 1
Nên (92)9058 cũng có chữ số tận cùng là 1
\(\Rightarrow\)918117 = (...1) .9
\(\Rightarrow\)918117 = (...9)
Vậy chữ số tận cùng của (99)2013 là 9
b) Ta có:
20081 = 2008 có chữ số tận cùng là 8
20082 = (...4) có chữ số tận cùng là 4 (vì 8.8=64)
20083 = (...2) có chữ số tận cùng là 2 (vì 4.8=32)
20084 = (...6) có chữ số tận cùng là 6 (vì 2.8=16)
và 2008100 = 20084.25= (20084)25
Vì 20084 có chữ số tận cùng là 6
Nên (20084)25 cũng có chữ số tận cùng là 6
Vậy 2008100 có chữ số tận cùng là 6
a: \(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{9}=\dfrac{11}{18}\)
hay \(x=\dfrac{11}{18}:\dfrac{1}{4}=\dfrac{11}{18}\cdot4=\dfrac{44}{18}=\dfrac{22}{9}\)
d: =>x+1;x-2 khác dấu
Trường hợp 1: \(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Leftrightarrow-1< x< 2\)
Trường hợp 2: \(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Leftrightarrow2< x< -1\left(loại\right)\)
e: =>x-2>0 hoặc x+2/3<0
=>x>2 hoặc x<-2/3
bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right)
\)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)
b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .
a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
... câu cuối bn lm dài dòng quá r ạ -)) cái dòng sra là bỏ luôn dấu GTTĐ của VT r ạ :))
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)