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英語 高校生

仮定法の範囲です 6〜21、20、24〜26の答え教えてください! 解いたのであってるかみてほしいです。不正解の場合正しい回答もしりたいです🙇‍♂️

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英語 高校生

教えてください!!

PAR Review 5 (Point 131~157) to smo r ■ QAZ- GREL 131~157) 形容詞の語法, 副詞の語法,比較 ETC 英文中の空所に入る適切な語または語句を選択肢から選びなさい。 2. There was hardly ( 1 more 1. Many students were not able to solve the problem, but ( ents wer 1 few ble to so 2 a few 3 little 3. Ken didn't work as ( 1 hard 3 hardest 4. He has a great ( things usd 1973 7. Your plan is ( 17. Ⓒ less do ) rainfall in that area of the country. 2 some 3 no 4 any slɔ od ai (-) ant eslood om hans and sil2 (dytas) ) were. O TOC Besides ) of knowledge about languages. sie o 2 deal 3 quality 形容詞の語法, 副詞の語法, 比較 viilsup ai ) as his brother, and he failed in the final exam. es dou 3 pleased *080 6. Ellie has to finish her project by next Monday. boss. 1 Except ) better than mine. 12 far to excite 5. The singers seemed a little too () with their success. be pleased 2 be pleasing July 11. The news was (\) to us. 21. Ⓒsurprise 2 more seriously 4 many serious 2 excitement PILN 2 surprised on bed anellob svit 4 a little 3 more 10. The school is strict about hair styles; ( (1 moreover 2 otherwise 3 Additionally (B2), 4 ) (0) die Gl and number gaigbulafe 3 surprising TOXOX bubong eidT) Vino 4 too 4 Otherwise 131 (東海大) um coM as 146 blaw edi buona 〈 名古屋学院大 > Har \ gad) 本日132 4 pleasing 〈 近畿大 > 185 she will get in trouble with her 8. A friend of mine is coming to Japan next month. I am very ( ) about the news. ➡132 1 excite (3) excited 4 exciting <東海大 > LOW LAW 304 3 exciting 4 excitable 1940 inging yM AUTO ), some students break the rules. 3 nevertheless 4 notwithstanding ro sniqgsH 9. By the time it ended, our team had battled hard and finally won. The game was indeed ( ). 10 excited 142 ➡131 <杏林大〉 ② 次の英文の 1. Count day. 4 to surprise JanoviaU bolx0 144 < 南山大 > 150 1) I bsando gli om man soad sms I asdW <玉川大〉 132 〈北里大〉 145 〈法政大> 132 <京都学園大 > REST ③ 次の日本 01. 1 Sh 2. Of 3. Alice □ 2. そ C ☐ 3. □ 4.

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数学 高校生

!!!至急お願いします!!! マーカーのところで、この式はどこから求めた式ですか?

576 基本例題 126 連立漸化式 (2) 数列{an},{bn} を α = 1, b1 = -1, an+1=5an-4bn, bn+1=an+bnで定めると (1)an+1+xbn+1=y(an+xbn) を満たすx, yの値を求めよ。 (2) 数列{an}, {bn}の一般項を求めよ。 MERA 指針 p.575 基本例題125 (1) と同様に, 〔解法1] 「等比数列を利用」 の方針によって解けばよ an+xbn=(a₁+xb₁) (2) (1) から,数列{an+xbn}は公比yの等比数列となり これに αn=bn+1-6 を代入し, an を消去すると bn+1=(1-x)bn+(a+xbi)yn-1 解答 (1) an+1+xbn+1=5an-4bn+x(an+bn) $27 an+1= pan+g 型の漸化式 (p.564 基本例題118) に帰着。 よって, ① の両辺をy7+1で割ればよい。 =(5+x)an+(-4+x)bn よって, an+1+xbn+1=y(an+xbn) とすると (5+x)an+(-4+x)bn=yan+xybn REOC I) (S これがすべてのnについて成り立つための条件は ...... 5+x=y, -4+x=xy 5+x=yを4+x=xy に代入して整理すると x2+4x+4=0 ゆえに したがって 求める x, yの値は (2) (1)から an+1-2bn+1=3(an-2bn) よって, 数列{an-26m}は,初項α1-261=3,公比3の等比 数列であるから x=-2 x=-2, y=3 bn+1 bn 1 + 3n+1 3" 3 + + an-26m=3.3"-1 3 すなわち an=26+3" これに an=bn+1- 6 を代入すると bn+1=36n+3 3¹ bn あるから 第1-131+(n-1)-13-032 = よって a=3"-¹(2n-1), b=3"-¹(n-2) 両辺を 37+1 で割ると 数列{10} は,初項 1/14-11/11/13 公差 1/13の等差数列で = ま め [参考] [解法2][1つの に関する漸化式に帰着させ る]の方針による解答 an+1=5an-4bn. bn+1=an+bn ② から a=bx+1-bm an+1=bn+2-b₁ これらを①に代入して bn+2-6bn+1+9bn=0 特性方程式x^2-6x+9=0 解くとx=3(重解) よって, p.573 基本例題124 と同じ方針で、 まず一般 を求める。 1 lan+1=pantg”型は両辺 g" +1 で割る(p.564 参照)。 an=26+3に代入。

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