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Bài 10:
1: \(\left(7-\frac15+\frac13\right)-\left(6+\frac95+\frac43\right)\)
\(=7-\frac15+\frac13-6-\frac95-\frac43\)
\(=\left(7-6\right)+\left(-\frac15-\frac95\right)+\left(\frac13-\frac43\right)\)
=1-2-1
=-2
2: \(7+\left(\frac{7}{12}-\frac12+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
=10-5
=5
3: \(\left(\frac12-\frac13\right)-\left(\frac53-\frac32\right)+\left(\frac73-\frac52\right)\)
\(=\frac12-\frac13-\frac53+\frac32+\frac73-\frac52\)
\(=-\frac12+\frac13=\frac{-3+2}{6}=-\frac16\)
4: \(\left(\frac27-\frac94\right)-\left(-\frac37+\frac54\right)-\left(\frac24-\frac97\right)\)
\(=\frac27-\frac94+\frac37-\frac54-\frac24+\frac97\)
\(=\left(\frac27+\frac37+\frac97\right)+\left(-\frac94-\frac54-\frac24\right)=\frac{14}{7}-\frac{16}{4}=2-4=-2\)
5: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)
\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)
\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)
\(=\frac33+\left(-3\right)=1-3=-2\)
Bài 11:
1: \(\frac25\cdot\frac38_{}+\frac58\cdot\frac25=\frac25\left(\frac38+\frac58\right)=\frac25\cdot\frac88=\frac25\)
2: \(\frac23\cdot\frac52-\frac34\cdot\frac23=\frac23\left(\frac52-\frac34\right)=\frac23\cdot\frac74=\frac{14}{12}=\frac76\)
3: \(\frac57\cdot\frac{19}{23}-\frac{12}{23}\cdot\frac57=\frac57\left(\frac{19}{23}-\frac{12}{23}\right)=\frac57\cdot\frac{7}{23}=\frac{5}{23}\)
4: \(\frac72\cdot\frac{11}{6}-\frac72\cdot\frac56=\frac72\left(\frac{11}{6}-\frac56\right)=\frac72\cdot\frac66=\frac72\)
5: \(\frac{11}{9}\cdot\frac34-\frac29\cdot\frac34=\frac34\left(\frac{11}{9}-\frac29\right)=\frac34\cdot\frac99=\frac34\)
6: \(\frac37\cdot\frac{13}{5}+\frac37\cdot\frac85=\frac37\left(\frac{13}{5}+\frac85\right)=\frac37\cdot\frac{21}{5}=\frac{21}{7}\cdot\frac35=3\cdot\frac35=\frac95\)
7: \(\frac{7}{15}\cdot\frac{16}{13}+\frac{7}{15}\cdot\frac{-3}{13}=\frac{7}{15}\left(\frac{16}{13}-\frac{3}{13}\right)=\frac{7}{15}\cdot\frac{13}{13}=\frac{7}{15}\)
8: \(-\frac{23}{7}\cdot\frac{3}{10}+\frac{13}{7}\cdot\frac{3}{10}=\frac{3}{10}\left(-\frac{23}{7}+\frac{13}{7}\right)=\frac{3}{10}\cdot\frac{-10}{7}=-\frac37\)
9: \(\frac{-11}{8}\cdot\frac{19}{3}+\frac{19}{3}\cdot\frac{-5}{8}=\frac{19}{3}\left(-\frac{11}{8}-\frac58\right)=\frac{19}{3}\cdot\left(-2\right)=-\frac{38}{3}\)
Bài 12: Bài 12:
1: \(\frac{-5}{17}\cdot\frac{31}{33}+\frac{-5}{17}\cdot\frac{2}{33}+1\frac{5}{17}\)
\(=-\frac{5}{17}\cdot\left(\frac{31}{33}+\frac{2}{33}\right)+1+\frac{5}{17}\)
\(=-\frac{5}{17}+1+\frac{5}{17}=1\)
2: \(\frac57\cdot\left(-\frac{3}{11}\right)+\frac57\cdot\left(-\frac{8}{11}\right)+2\frac57\)
\(=-\frac57\left(\frac{3}{11}+\frac{8}{11}\right)+2+\frac57\)
\(=-\frac57+2+\frac57=2\)
3: \(\frac{9}{10}\cdot\frac{23}{11}-\frac{1}{11}\cdot\frac{9}{10}+\frac{9}{10}\)
\(=\frac{9}{10}\left(\frac{23}{11}-\frac{1}{11}+1\right)\)
\(=\frac{9}{10}\cdot\left(2+1\right)=\frac{9}{10}\cdot3=\frac{27}{10}\)
4: \(\frac54\cdot\frac{8}{15}+\frac{-5}{16}\cdot\frac{8}{15}-1\)
\(=\frac{8}{15}\left(\frac54-\frac{5}{16}\right)-1\)
\(=\frac{8}{15}\left(\frac{20}{16}-\frac{5}{16}\right)-1=\frac{8}{16}-1=-\frac{8}{16}=-\frac12\)
5: \(-\frac{19}{3}\cdot\frac{14}{4}+\frac{25}{4}\cdot\frac{-19}{3}+4\frac34\)
\(=-\frac{19}{4}\left(\frac{14}{3}+\frac{25}{3}\right)+4\frac34\)
\(=-\frac{19}{4}\cdot13+\frac{19}{4}=\frac{19}{4}\left(-13+1\right)=\frac{19}{4}\cdot\left(-12\right)=-57\)
6: \(\frac{1}{27}\cdot\frac{-3}{7}-\frac59\cdot\frac{-3}{7}+\frac19\)
\(=-\frac37\left(\frac{1}{27}-\frac59\right)+\frac19\)
\(=-\frac37\left(\frac{1}{27}-\frac{15}{27}\right)+\frac19=-\frac37\cdot\frac{-14}{27}+\frac19=\frac29+\frac19=\frac39=\frac13\) b

Bài 14:
\(A\left(x\right)+B\left(x\right)=5x^4-6x^3-3x^2-4\)
\(A\left(x\right)-B\left(x\right)=3x^4+7x^2+8x+2\)
Do đó: \(A\left(x\right)+B\left(x\right)+A\left(x\right)-B\left(x\right)=5x^4-6x^3-3x^2-4+3x^4+7x^2+8x+2\)
=>\(2\cdot A\left(x\right)=8x^4-6x^3+4x^2+8x-2\)
=>\(A\left(x\right)=4x^4-3x^3+2x^2+4x-1\)
Ta có: \(A\left(x\right)+B\left(x\right)=5x^4-6x^3-3x^2-4\)
=>\(B\left(x\right)=5x^4-6x^3-3x^2-4-4x^4+3x^3-2x^2-4x-1\)
=>\(B\left(x\right)=x^4-3x^3-5x^2-4x-5\)
Bài 13:
\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\)
\(f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\)
Do đó: \(f\left(x\right)+g\left(x\right)+f\left(x\right)-g\left(x\right)=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\)
=>\(2\cdot f\left(x\right)=10x^4-6x^3+4x^2+8x-14\)
=>\(f\left(x\right)=5x^4-3x^3+2x^2+4x-7\)
\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\)
=>\(g\left(x\right)=6x^4-3x^2-5-5x^4+3x^3-2x^2-4x+7=x^4+3x^3-5x^2-4x+2\)

bài 2: a. ta có góc ADE = góc ABC (= 45 độ)
mà 2 góc này ở vị trí đồng vị
⇒ DE // BC
b. ta có góc FEC = góc ECB
mà 2 góc này ở vị trí so le trong
⇒ EF // BC
c. vì DE // BC và EF // BC nên DE ≡ EF
⇒ 3 điểm D,E,F thẳng hàng
bài 3:
a. ta có góc CHK = góc CAB = 90 độ
mà 2 góc này ở vị trí đồng vị
⇒ KH // AB
b. ta có góc IKB = góc KBA = 60 độ
mà 2 góc này ở vị trí so le trong
⇒ KI // AB
c. vì KH // AB và KI // AB nên KH ≡ KI
⇒ 3 điểm H,K,I thẳng hàng

4.
\(\left(0,36\right)^8=\left(\left(0,6\right)^2\right)^8=\left(0,6\right)^{16}\)
\(\left(0,216\right)^4=\left(\left(0,6\right)^3\right)^4=\left(0,6\right)^{12}\)
5.
a, \(\left(3\times5\right)^3=15^3=1125\)
b, \(\left(\frac{-4}{11}\right)^2=\frac{16}{121}\)
c, \(\left(0,5\right)^4\times6^4=\left(0,5\times6\right)^4=3^4=81\)
d, \(\left(\frac{-1}{3}\right)^5\div\left(\frac{1}{6}\right)^5=\left(\frac{-1}{3}\right)^5\times6^5=\left(\frac{-1}{3}\times6\right)^5=\left(-2\right)^5=-32\)
6.
a, \(\frac{6^2\times6^3}{3^5}=\frac{6^5}{3^5}=\frac{2^5\times3^5}{3^5}=2^5=32\)
b, \(\frac{25^2\times4^2}{5^5\times\left(-2\right)^5}=\frac{100^2}{\left(-10\right)^5}=\frac{10^4}{\left(-10\right)^5}=\frac{-1}{10}\)
c, Mình không nhìn rõ đề
d, \(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2=\left(\frac{-11}{4}+\frac{1}{2}\right)^2=\left(\frac{-9}{4}\right)^2=\frac{81}{16}\)
7.
a, \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\Rightarrow m=4\)
b, \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\Rightarrow\left(\frac{3}{5}\right)^n=\left(\left(\frac{3}{5}\right)^2\right)^5\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\Rightarrow n=10\)
c, \(\left(-0,25\right)^p=\frac{1}{256}\Rightarrow\left(-0,25\right)^p=\left(\frac{1}{4}\right)^4\Rightarrow\left(-0,25\right)^p=\left(0,25\right)^4\Rightarrow p=4\)
8.
a, \(\left(\frac{2}{5}+\frac{3}{4}\right)^2=\left(\frac{23}{20}\right)^2=\frac{529}{400}\)
b, \(\left(\frac{5}{4}-\frac{1}{6}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d: ĐKXĐ: x>=2
Ta có: \(\left(3\sqrt{x-2}+2\right)\left(\sqrt{x-1}+x\right)=0\)
mà \(3\sqrt{x-2}+2\ge2>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x-1}=x\)
=>\(\begin{cases}x-1=x^2\\ x\ge0\end{cases}\Rightarrow\begin{cases}x^2-x+1=0\\ x\ge2\end{cases}\)
=>\(\begin{cases}x^2-x+\frac14+\frac34=0\\ x\ge2\end{cases}\Rightarrow\begin{cases}\left(x-\frac12\right)^2+\frac34=0\left(vôlý\right)\\ x\ge2\end{cases}\)
=>x∈∅

1:gia tri x<0...
2:gia tri x thoa man...
3:gia tri a biet...
4:-2,1
5:26/64...
6:gia tri bieu thuc (2/5)7...
7:1-2/3...
8:4 va 3/4
10:gia tri bieu thuc :24+....
nếu không phải thì bạn đổi 8 rồi tới 7 nhé !!!!!!

Bài 4:
a: Xét ΔADB và ΔADC có
AB=AC
\(\widehat{BAD}=\widehat{CAD}\)
AD chung
Do đó: ΔADB=ΔADC
b: Xét ΔAHD vuông tại H và ΔAKD vuông tại K có
AD chung
\(\widehat{HAD}=\widehat{KAD}\)
Do đó: ΔAHD=ΔAKD
Suy ra: AH=AK
c: Đặt \(\widehat{A}=a;\widehat{C}=c\)
Theo đề, ta có: \(\left\{{}\begin{matrix}a=3c\\a=180-2c\end{matrix}\right.\Leftrightarrow3c=180-2c\)
=>c=36
=>\(\widehat{ACB}=\widehat{ABC}=36^0\)
=>\(\widehat{BAC}=108^0\)
là sao , mk ko hiểu