tìm x:
\(x-\frac{2}{8}=\frac{8}{4}\)
Những câu hỏi liên quan
cho x,y,z là số thực ,\(xyz=2\sqrt{2}\)
Tìm GTNN của \(P=\frac{x^8+y^8}{x^4+y^4+x^2y^2}+\frac{x^8+z^8}{x^4+z^4+x^2z^2}+\frac{y^8+z^8}{y^4+z^4+y^2z^2}\)
1) \(\frac{24}{-12}=\frac{x}{5}=\frac{-y}{3}\)Tìm x và y
2) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-5}{25}\le\frac{x}{10}< \frac{-3}{4}+\frac{4}{14}+\frac{-2}{8}+\frac{-3}{5}+\frac{5}{7}\)Tìm x
3) \(\frac{8.x+18}{2.x+6}\)Tìm x
Tìm x ; y Biết
\(\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4+y^4}+\frac{8y^8}{x^8-y^8}=4\)
Và x + y = - 9
Ai tick mik vài cái cho tròn 170 với
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\(\frac{2}{x-1.x-3}+\frac{5}{x-3.x-8}+\frac{12}{x-8.x-20}-\frac{1}{x-20}=\frac{-3}{4}\)
tìm x
Có lẽ bạn viết đề sai.
Câu hỏi của Vũ Mai Linh - Toán lớp 7 - Học toán với OnlineMath
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tìm x: \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{\left(x+16\right)-\left(x+2\right)}{\left(x+2\right)\left(x+16\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow x+16-x-2=x\)
\(\Rightarrow x=14\)
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Tìm x,y biết :
a.\(\frac{x^2}{4}=\frac{y}{2}\)và x2 - 3y = -8
b. \(\frac{x+1}{8}=\frac{8}{x+1}\)( x khác 1)
Áp dụng tính chất dãy tỉ số bằng nhau
ta có ; \(\frac{x^2}{4}=\frac{y}{2}=\frac{x^2-3y}{4-\left(3.2\right)}=\frac{-8}{-2}=4.\)
\(\Rightarrow\frac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x=\pm4\)
\(\Rightarrow\frac{y}{2}=4\Rightarrow y=8\)
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b) \(\frac{x+1}{8}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=64=8^2=\left(-8\right)^2\)
=> x + 1 = 8 => x = 7
x+1 = -8 => x = -9
KL:...
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Tìm x biết:
a)\(\frac{2}{3}.\left(x-\frac{3}{8}\right)-x-\left(-\frac{7}{8}+\frac{2}{3}\right)=\left(\frac{-3}{4}\right)^3:1\frac{11}{16}\)
b)\(-\frac{7}{8}+\frac{7}{8}:\left(\frac{2}{3}-x\right)+\frac{5}{6}:\left(-1\frac{11}{35}\right)=\left(0,8\right)^2\)
Tìm x biết \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+4\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
Tìm x thuộc Q biết: \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
ĐKXĐ:\(x\ne\left\{-2;-4;-8;-14\right\}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow2\left(x+8\right)\left(x+14\right)+4\left(x+2\right)\left(x+14\right)+6\left(x+2\right)\left(x+4\right)=x\left(x+8\right)\left(x+14\right)\)
\(\Leftrightarrow2x^2+44x+224+4x^2+64x+112+6x^2+36x+48=x^3+22x^2+112x\)
\(\Leftrightarrow12x^2+144x+384=x^3+22x^2+112x\)
\(\Leftrightarrow x^3+22x^2-12x^2+112x-144x-384=0\)
\(\Leftrightarrow x^3+10x^2-32x-384=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2+16x+64\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+8\right)^2=0\)
\(\Leftrightarrow x=6\)(x=-8 loại vì x=-8 thì PT không xác định)
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Điều kiện: x+ 2 \(\ne\) 0 ; x+ 4 \(\ne\) 0; x+ 8 \(\ne\) 0 ; x + 14 \(\ne\) 0
<=> \(\frac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\frac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\frac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
<=> \(\frac{x+4}{\left(x+2\right)\left(x+4\right)}-\frac{x+2}{\left(x+2\right)\left(x+4\right)}+\frac{x+8}{\left(x+4\right)\left(x+8\right)}-\frac{x+4}{\left(x+4\right)\left(x+8\right)}+\frac{x+14}{\left(x+8\right)\left(x+14\right)}-\frac{x+8}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)<=> \(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
<=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
<=> \(\frac{12\left(x+4\right)}{\left(x+2\right)\left(x+14\right)\left(x+4\right)}=\frac{x\left(x+14\right)}{\left(x+2\right)\left(x+4\right)\left(x+14\right)}\)
<=> 12(x + 4) = x (x + 14)
<=> 12x + 48 = x2 + 14 x
<=> x2 + 2x - 48 = 0
<=> x2 + 8x - 6x - 48 = 0
<=> x(x + 8) - 6 (x + 8) = 0
<=> (x - 6)(x + 8) = 0 <=> x - 6 = 0 (do x + 8 \(\ne\) 0)
<=> x = 6
Vậy x = 6
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