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Ta có: \(\frac{x}{y}=\frac{5}{4}\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\Rightarrow\frac{2x^2}{50}=\frac{3y^2}{48}\) (*)
Từ (*),áp dụng t/c tỉ dãy số bằng nhau,ta có:
\(\frac{2x^2}{50}=\frac{3y^2}{48}=\frac{2x^2-3y^2}{50-48}=\frac{288}{2}=144\)
\(\Rightarrow\hept{\begin{cases}2x^2=7200\\3y^2=6912\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=3600\\y^2=2304\end{cases}}\Leftrightarrow\hept{\begin{cases}x=60\\y=48\end{cases}}}\)
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a. 2x + 2x+3 = 288
<=> 2x.(1+23) = 288
<=> 2x = 32
<=> 2x = 25
<=> x = 5
b. (2x - 1)3 = 27
<=> (2x - 1)3 = 33
<=> 2x - 1 = 3
<=> x = 2
c. (x + 5)3 = -64
<=> (x + 5)3 = (-4)3
<=> x + 5 = -4
<=> x = -9
d. (2x - 3)2 = 9
<=> (2x - 3)2 = 32
<=> 2x - 3 = 3
<=> x = 3
VÀO TRANG CÁ NHÂN CỦA E
Em bức xúc lắm anh chị ạ, xl mấy anh chị vì đã gây rối
Thiệt tình là ko chấp nhận nổi con nít ms 2k6 mà đã là vk là ck r ạ, bày đặt yêu xa, chưa lên đại học
Đây là 'tội nhân'
https://olm.vn/thanhvien/nhu140826 và https://olm.vn/thanhvien/trungkienhy79
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Ta có : 5x = 2y \(\Rightarrow15x=6y\)
3y = 5z \(\Rightarrow6y=10z\)
Nên : \(\Rightarrow15x=6y=10z\)
Ta có ; \(15x=6y=10z=\frac{x}{\frac{1}{15}}=\frac{y}{\frac{1}{6}}=\frac{z}{\frac{1}{10}}=\frac{2x}{\frac{2}{15}}=\frac{3y}{\frac{3}{6}}=\frac{2x-3y+z}{\frac{2}{15}-\frac{3}{6}+\frac{1}{10}}=\frac{288}{-\frac{4}{15}}=-1080\)
Nên 15x = -1080 => x = -72
6y = -1080 => y = -180
10z = -1080 => z = -108
Vậy x = -72 ; y = -180 ; z = -108
\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{5}\);\(\frac{y}{5}=\frac{z}{3}\)
\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{5}=\frac{z}{3}\)
\(\Rightarrow\)\(\frac{2x}{4}=\frac{x}{2}\);\(\frac{3y}{15}=\frac{y}{5}\);\(\frac{z}{3}\)
\(\Rightarrow\)\(\frac{2x}{4}=\frac{3y}{15}=\frac{z}{3}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{2x-3y+z}{4-15+3}=\frac{288}{-8}=-36\)
\(\Rightarrow\)\(\frac{x}{2}=-36\Rightarrow x=-72\)
\(\frac{y}{5}=-36\Rightarrow y=-180\)
\(\frac{z}{3}=-36\Rightarrow z=-108\)
Vậy x , y, z lần lượt là: -72, -180, -108.
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a.\(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\)
\(\Rightarrow\left(-\frac{1}{6}\right)^{x-1}=\frac{1}{36}\)
\(\Rightarrow \left(-\frac{1}{6}\right)^{x-1}=\left(-\frac{1}{6}\right)^2\)
=> x-1=2
=> x=2+1
Vậy x=3.
b.\(81^{-2x}.27^x=9^5\)
\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=\left(3^2\right)^5\)
\(\Rightarrow3^{4.\left(-2x\right)}.3^{3x}=3^{10}\)
\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)
\(\Rightarrow3^{-5x}=3^{10}\)
=> -5x=10
=> x=10:(-5)
Vậy x=-2.
c.\(2^x+2^{x+3}=288\)
\(\Rightarrow2^x.\left(1+2^3\right)=288\)
\(\Rightarrow2^x.9=288\)
\(\Rightarrow2^x=288:9\)
\(\Rightarrow2^x=32\)
=> 2x=25
Vậy x=5.
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a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
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a, \(4.3^{2x}-2.9^x-54=0\)
\(\Rightarrow3^{2x}\left(4-2\right)=54\)
\(\Rightarrow3^{2x}=27=3^3\)
Vì \(3\ne\pm1;3\ne0\) nên \(2x=3\Rightarrow x=\dfrac{3}{2}\)
b, \(\dfrac{1}{2}.2^x+4.2^x-288=0\)
\(\Rightarrow2^x\left(\dfrac{1}{2}+4\right)=288\)
\(\Rightarrow2^x=64=2^6\)
Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)
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a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)
Bậc của P(x) là 3
\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)
Bậc của Q(x) là 3
b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)
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a) \(\left|2x-2\right|-2x=3\)
\(\Rightarrow\left|2x+2\right|=3+2x\)
\(\Rightarrow2x+2=\pm\left(3+2x\right)\)
+) \(2x+2=3+2x\Rightarrow2=3\) ( không thỏa mãn )
+) \(2x+2=-\left(3+2x\right)\)
\(\Rightarrow2x+2=3-2x\)
\(\Rightarrow2x+2x=3-2\)
\(\Rightarrow4x=1\)
\(\Rightarrow x=\frac{1}{4}\) ( thỏa mãn )
Vậy \(x=\frac{1}{4}\)
b) \(\left|2x+3\right|+2x=-3\)
\(\Rightarrow\left|2x+3\right|=-3-2x\)
\(\Rightarrow2x+3=\pm\left(-3-2x\right)\)
+) \(2x+3=-3-2x\)
\(\Rightarrow2x+2x=-3-3\)
\(\Rightarrow4x=-6\)
\(\Rightarrow x=\frac{-3}{2}\) ( thỏa mãn )
+) \(2x+3=-\left(-3-2x\right)\)
\(\Rightarrow2x+3=-3+2x\)
\(\Rightarrow3=-3\) ( không thỏa mãn )
Vậy \(x=\frac{-3}{2}\)
=>2x.(1+23)=288
=>2x.9=288
=>2x=288:9
=>2x=32
=>x=5
2x+2x+3=288
2x.1+2x.23=288
2x(1+23)=288
2x.9=288
2x=288:9
2x=32
2x=25
=> x=5
Vậy x=5