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a) Ta có : \(\frac{2}{3}x=\frac{3}{4}y=\frac{5}{6}z\)=> \(\frac{2x}{3}=\frac{3y}{4}=\frac{5z}{6}\)=> \(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{6}{5}}\)
=> \(\frac{x^2}{\frac{9}{4}}=\frac{y^2}{\frac{16}{9}}=\frac{z^2}{\frac{36}{25}}\)
Đặt \(\frac{x^2}{\frac{9}{4}}=\frac{y^2}{\frac{16}{9}}=\frac{z^2}{\frac{36}{25}}=k\Leftrightarrow\hept{\begin{cases}x^2=\frac{9}{4}k\\y^2=\frac{16}{9}k\\z^2=\frac{36}{25}k\end{cases}}\)
=> \(x^2+y^2+z^2=\frac{9}{4}k+\frac{16}{9}k+\frac{36}{25}k\)
=> \(\frac{4921}{900}k=724\)
=> \(k=724:\frac{4921}{900}=\frac{651600}{4921}\)
Do đó : \(\hept{\begin{cases}x^2=\frac{9}{4}\cdot\frac{651600}{4921}\\y^2=\frac{16}{9}\cdot\frac{651600}{4921}\\z^2=\frac{36}{25}\cdot\frac{651600}{4921}\end{cases}}\)
Bài toán đây có sai sót j không vậy?Thấy số dữ quá đi :v
b) Ta có : \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}\)
=> \(\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}=\frac{x-1-2y+4+3z-9}{2-6+12}=\frac{x-2y+3z-6}{8}=\frac{46-6}{8}=\frac{40}{8}=5\)
=> \(\hept{\begin{cases}\frac{x-1}{2}=5\\\frac{y+2}{3}=5\\\frac{z-3}{4}=5\end{cases}}\Rightarrow\hept{\begin{cases}x=11\\y=13\\z=23\end{cases}}\)
c) Đặt \(\frac{x}{3}=\frac{y}{16}=k\Rightarrow\hept{\begin{cases}x=3k\\y=16k\end{cases}}\)
=> xy = 16k . 3k
=> 48k2 = 192
=> k2 = 4
=> k = 2 hoặc k = -2
Do đó \(\left(x,y\right)\in\left\{\left(6,32\right);\left(-6,-32\right)\right\}\)
Bài 2 : a) \(\frac{4^2\cdot25^2+16\cdot125}{2^3\cdot5^2}\)
\(=\frac{\left(2^2\right)^2\cdot\left(5^2\right)^2+16\cdot125}{2^3\cdot5^2}\)
\(=\frac{2^4\cdot5^4+2^4\cdot5^3}{2^3\cdot5^2}\)
\(=\frac{2\cdot2^3\left(5^4+5^3\right)}{2^3\cdot5^2}\)
\(=\frac{2\cdot5^3\left(5+1\right)}{5^2}=\frac{2\cdot5\cdot5^2\cdot6}{5^2}=2\cdot5\cdot6=60\)
b) \(\frac{6^8\cdot2^4-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)
\(=\frac{\left(2\cdot3\right)^8\cdot2^4-\left(2^2\right)^5\cdot\left(2\cdot3^2\right)^4}{\left(3^3\right)^3\cdot\left(2^3\right)^4-3^9\cdot2^{13}}\)
\(=\frac{2^8\cdot3^8\cdot2^4-2^{10}\cdot2^4\cdot3^8}{3^9\cdot2^{12}-3^9\cdot2^{13}}\)
\(=\frac{2^{12}\cdot3^8-2^{14}\cdot3^8}{3^9\left(2^{12}-2^{13}\right)}\)
\(=\frac{3^8\left(2^{12}-2^{14}\right)}{3^9\left(2^{12}-2^{13}\right)}=\frac{3^8\left(2^{12}-2^{14}\right)}{3^8\left(2^{12}-2^{13}\right)\cdot3}=1\)
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)
\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)
\(=-\frac{1}{2}x^2y^2\)
2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)
\(=\frac{17}{6}x^2\)
3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)
\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)
\(=-\frac{67}{4}x^2y^3\)
4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)
\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)
\(=-\frac{97}{30}x^4y\)
5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)
\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)
\(=-\frac{5}{12}x^6y^8\)
bn ơi,vì tất cả bài tập này khá nhiều và cx khá khó nên sẽ ko ai trả lời đâu,bn nên đăng từng bài một thôi nhé,nếu bn đăng như mk nói thì mà ko có ai trả lời thì hãy viết bài toán đó trên google để tra nhé,chúc bn làm bài tốt
b)ta có: \(\frac{x}{5}=\frac{y}{4}=\frac{z}{-6}\Rightarrow\frac{x^3}{125}=\frac{y^3}{64}=\frac{z^3}{-216}=\frac{x^3}{125}=\frac{y^3}{64}=\frac{3z^3}{-648}\)
ADTCDTSBN
có: \(\frac{x^3}{125}=\frac{3z^3}{-648}=\frac{x^3+3z^3}{125+\left(-648\right)}=\frac{-14121}{-523}=27\)
=> x3/125 = 27 => x3 = 3 375 => x = 15
y3/64 = 27 => y3 = 1 728 => y = 12
z3/-216 =27 => z3 = -5 832 => z3 = -18
KL:...
câu c thì mk ko bk! sr bn nha!
a) ta có: \(\frac{x}{y}=\frac{7}{20}\Rightarrow x20=y7\Rightarrow\frac{x}{7}=\frac{y}{20}\Rightarrow\frac{x}{49}=\frac{y}{140}\)
\(\frac{y}{z}=\frac{7}{3}\Rightarrow y3=z7\Rightarrow\frac{y}{7}=\frac{z}{3}\Rightarrow\frac{y}{140}=\frac{z}{60}\)
\(\Rightarrow\frac{x}{49}=\frac{y}{140}=\frac{z}{60}\)
ADTCDTSBN
có: \(\frac{x}{49}=\frac{y}{140}=\frac{z}{60}=\frac{x-y+z}{49-140+60}=\frac{-155}{-31}=5\)
=> x/49 = 5 => x = 245
y/140 = 5 => y = 700
z/60 = 5 => z = 300
KL:...
1) Ta có: 5x = 2y = x/2 = y/5
Đặt \(\frac{x}{2}=\frac{y}{5}=k\) => \(\hept{\begin{cases}x=2k\\y=5k\end{cases}}\) (*)
Khi đó, ta có: x3y2 = 200
=> (2k)3.(5k)2 = 200
=> 8k3 . 25k2 = 200
=> 200k5 = 200
=> k5 = 1
=> k = 1
Thay k = 1 vào (*), ta được:
+) x = 2.1 = 2
+) y = 5.1 = 5
Vậy ...
1.
\((\frac{1}{3}xy)^2.x^3+\frac{3}{2}(2x)^3(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)
\(=(\frac{1}{9}x^2y^2)x^3+\frac{3}{2}(8x^3)(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)
\(=\frac{1}{9}(x^2.x^3)y^2+(\frac{3}{2}.8.\frac{-7}{4})(x^3.x^2).y^2-\frac{2}{3}x^5y^2\)
\(=\frac{1}{9}x^5y^2-21x^5y^2-\frac{2}{3}x^5y^2=\frac{-194}{9}x^5y^2\)
2.
\(\frac{-2}{5}x^2y(-y^6)+\frac{3}{2}xy(\frac{-1}{15}xy^6)+(-2xy)^2y^5\)
\(=\frac{2}{5}x^2(y.y^6)+(\frac{3}{2}.\frac{-1}{15})(x.x).(y.y^6)+4x^2(y^2.y^5)\)
\(=\frac{2}{5}x^2y^7-\frac{1}{10}x^2y^7+4x^2y^7=\frac{43}{10}x^2y^7\)
3.
\(\frac{3}{7}xy^2z+\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2-\frac{3}{7}xy^2z\)
\(=(\frac{3}{7}xy^2z-\frac{3}{7}xy^2z)+(\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2)\)
\(=\frac{5}{6}x^3y^2\)
4.
\(\frac{2}{3}xy^2-\frac{5}{2}yz+\frac{1}{2}xy^2-\frac{2}{3}yz\)
\(=(\frac{2}{3}xy^2+\frac{1}{2}xy^2)-(\frac{5}{2}yz+\frac{2}{3}yz)\)
\(=\frac{7}{6}xy^2+\frac{19}{6}yz\)
5.
\(\frac{3}{2}xy^2z^5-\frac{5}{4}xyz^2+\frac{4}{3}xy^2z^5+\frac{1}{2}xyz^2\)
\(=(\frac{3}{2}xy^2z^5+\frac{4}{3}xy^2z^5)+(\frac{-5}{4}xyz^2+\frac{1}{2}xyz^2)\)
\(=\frac{17}{6}xy^2z^5-\frac{3}{4}xyz^2\)
\(\frac{x}{3}=\frac{y}{4}\)
=> \(\frac{x^2}{3^2}=\frac{y^2}{4^2}=\frac{x^2+y^2}{3^2+4^2}=\frac{400}{25}=16\)
=> x2 = 16 . 32 = 144 => x = 12 hoặc x = -12
y2 = 16 . 42 = 256 => y = 16 hoặc y = -16
KL: (x; y) = (12; 16) ; (-12; -16)