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=>\(\frac{x^2}{5^2}=\frac{y^2}{4^2}\)
=>\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{36}{9}=4\) (theo t/c của dãy TSBN)
=>\(\left(\frac{x}{5}\right)^2=\left(\frac{y}{4}\right)^2=2^2=\left(-2\right)^2\)
TH1:\(\frac{x}{5}=\frac{y}{4}=2\)
=>x=2.5=10 và y=2.4=8
TH2:\(\frac{x}{5}=\frac{y}{4}=-2\)
=>x=-2.5=-10 và y=-2.4=-8
Vậy (x;y) E {(10;8);(-10;-8)}
\(\frac{x}{5}=\frac{y}{4}\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{16}\)
Áp dụng tc dãy tỉ
\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{36}{9}=4\)
Với \(\frac{x^2}{25}=4\Rightarrow x=10\)
Với \(\frac{y^2}{16}=4\Rightarrow y=8\)
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1, 27x.9x=927:81
(33)x.(32)X=(32)27:34
33X.32X=354:34
33X+2X=350
35X=350
Suy ra :5x=50
x=10
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(5x + 1)2 = 36/49
=> (5x + 1)2 = (6/7)2
=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)
Làm từ phần b nha
b) \(\left(x-\frac{1}{9}\right)^3=\frac{2}{3}^6\)
\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{1}{3}\right)^6\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1^6}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{729}\)
\(\Rightarrow x-\frac{2}{9}=\frac{1}{9}\)
\(x=\frac{1}{9}+\frac{2}{9}\)
\(x=\frac{3}{9}=\frac{1}{3}\)
c) Sai đề rồi, xem lại đi
d) \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4< 0\)
\(\Rightarrow\frac{10000y^4-4000y^3+600y^3-40y+10000x^2+122501-70000x}{10000}< 0\)
=> Sai \(\forall y\inℝ\)
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\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x-7=\pm1\end{cases}}}\)
vậy x=7, x=8 hay x=6
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- a.(3x)2=1/243x33=1/9
3x=1/3 hoặc 3x=-1/3 ( vế 2 ko có x thỏa mãn)
suy ra x=3-1
b.(5x+1)=\(\sqrt{\frac{36}{49}}\)\(\Rightarrow\)5x+1=\(\frac{4}{7}\)hoặc 5x+1=\(\frac{-4}{7}\) | |
\(\Rightarrow\)x=\(\frac{-3}{35}\)hoặc x=\(\frac{-11}{35}\) | |
c.\(\frac{6}{4}\)-10x = \(\frac{4}{5}\)-3x chuyển vế :\(\frac{6}{4}\)-\(\frac{4}{5}\)= -3x + 10x \(\frac{7}{10}\)=7x \(\Rightarrow\)x =\(\frac{7}{10}\):7 \(\Rightarrow\)x= \(\frac{1}{10}\) |
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a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(\Leftrightarrow5x=\frac{-1}{7}\)
\(\Leftrightarrow x=\frac{-1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^2=\left(\frac{8}{27}\right)^2\)
đến đây làm giống câu a)
5(x - 4)2 - 9 = 36
5(x - 4)2 = 45
(x - 4)2 = 9
\(\Rightarrow\) x - 4 = 3 hoặc x - 4 = - 3
\(\Rightarrow\) x = 7 hoặc x = 1
\(5\left(x-4\right)^2-9=36\)
\(5\left(x-4\right)^2=36+9=45\)
\(\left(x-4\right)^2=45:5=9\)
\(\left(x-4\right)^2=3^3=\left(-3\right)^2\)
Trường hợp 1: \(x-4=3\Leftrightarrow x=3+4=7\)
Trường hợp 2: \(x-4=-3\Leftrightarrow x=-3+4=1\)
Vậy x = 7 hoặc x = 1