\(-x-\frac{2}{3}=-\frac{6}{7}\) giúp em gấp với ạ
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f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
h.3x - 2/6 - 5 = 3 - 2(x + 7)/4
<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4
<=> 3x - 32/6 = 3 - 2x - 14/4
<=> 3x - 32/6 = -2x - 11/4
<=> 6x - 64/12 = -6x - 33/12
<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12
Mình đề câu a phải như vậy nè:
\(a,\hept{\begin{cases}\frac{1}{x-2}+\frac{1}{y-1}=1\\\frac{2}{x-2}-\frac{3}{y-1}=1\end{cases}}\)\(Đkxđ:\hept{\begin{cases}x\ne2\\y\ne1\end{cases}}\)
Đặt: \(X=\frac{1}{x-2};Y=\frac{1}{y-1}\)
Ta có hệ sau:
\(\hept{\begin{cases}X+Y=1\\2X-3Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2\left(1-Y\right)-3Y=1\end{cases}}}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2-5Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=\frac{4}{5}\\Y=\frac{1}{5}\end{cases}}}\)
Với \(X=\frac{4}{5}\Rightarrow\frac{1}{x-2}=\frac{4}{5}\Leftrightarrow4\left(x-2\right)=5\Leftrightarrow x=\frac{13}{4}\)
Với \(Y=\frac{1}{5}\Rightarrow\frac{1}{y-1}=\frac{1}{5}\Leftrightarrow y-1=5\Leftrightarrow y=6\)
Vậy nghiệm của hệ pt là: \(\left(x;y\right)=\left(\frac{13}{4};6\right)\)
Câu b e nghĩ đề như vậy nè:
\(b,\hept{\begin{cases}\frac{7}{\sqrt{x-7}}-\frac{4}{\sqrt{y+6}}=\frac{5}{3}\\\frac{5}{\sqrt{x-7}}+\frac{3}{\sqrt{y+6}}=\frac{3}{6}\end{cases}}\) \(Đkxđ:\hept{\begin{cases}x>7\\x>-6\end{cases}}\)
Đặt \(\frac{1}{\sqrt{x-7}}=a\left(a>0\right);\frac{1}{\sqrt{y+6}}=b\left(b>0\right)\)
Ta có hệ pt mới: \(\hept{\begin{cases}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\end{cases}}\left(tmđk\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x-7=9\\x+6=36\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16\\y=30\end{cases}\left(tmđk\right)}\)
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(16;30\right)\)
(2/3×x-1/3)=2/3+1/3
(2/3×x-1/3)=3/3
2/3×x=3/3+1/3
2/3×x=4/3
x=4/3:3/2
x=4/3×2/3
x=8/9
\(M=\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{17}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
\(M=\frac{\left(\frac{1}{30}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{24}{119}+\frac{3}{35}\right).\frac{-4}{3}}\)
\(M=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{171}{595}.\frac{-4}{3}}\)
\(M=\frac{-1}{12}:\frac{-228}{595}\)
\(M=\frac{595}{2736}\)
Ta có:
\(M=\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right)\times\frac{5}{19}}{\left(\frac{1}{17}+\frac{1}{7}-\frac{-3}{35}\right)\times\frac{-4}{3}}\)
\(M=\frac{\left(\frac{1}{30}-\frac{7}{20}\right)\times\frac{5}{19}}{\left(\frac{24}{119}+\frac{3}{35}\right)\times\frac{-4}{3}}\)
\(M=\frac{\frac{-19}{60}\times\frac{5}{19}}{\frac{171}{595}\times\frac{-4}{3}}\)
\(M=\frac{-1}{12}\div\frac{-228}{595}\)
\(M=\frac{595}{2736}\)
Vậy \(M=\frac{595}{2736}\)
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
ĐK: x >0
Liên hợp:
pt <=> \(\sqrt{\frac{x^2+3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
<=> \(\frac{\frac{x^2+3}{x}-4}{\sqrt{\frac{x^2+3}{x}}+2}=\frac{x^2+7-4\left(x+1\right)}{2\left(x+1\right)}\)
<=> \(\frac{x^2-4x+3}{x\left(\sqrt{\frac{x^2+3}{x}}+2\right)}=\frac{x^2-4x+3}{2\left(x+1\right)}\)
<=> \(\orbr{\begin{cases}x^2-4x+3=0\left(1\right)\\x\left(\sqrt{\frac{x^2+3}{x}}+2\right)=2\left(x+1\right)\left(2\right)\end{cases}}\)
(1) <=> x = 1 hoặc x = 3 (tm)
(2) <=> \(x\sqrt{\frac{x^2+3}{x}}=2\)
<=> \(x\left(x^2+3\right)=4\)
<=> \(x^3+3x-4=0\)
,<=> (x-1)(x^2 +x +4) = 0
<=> x = 1 (tm)
Vậy x = 1 hoặc x = 3.
cách khác nhung chỉ dài thêm thôi
\(DK:x>0\)
PT\(\Leftrightarrow2\left(x+1\right)\sqrt{x^2+3}=\sqrt{x}\left(x^2+7\right)\)
Dat \(\sqrt{x^2+3}=t>0\)
PT tro thanh
\(\sqrt{x}t^2-2\left(x+1\right)t+4\sqrt{x}=0\)
Ta co:
\(\Delta^`_t=\left(x-2\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}t_1=\frac{x+1+\left|x-2\right|}{\sqrt{x}}\\t_2=\frac{x+1-\left|x-2\right|}{\sqrt{x}}\\t_3=\frac{x+1}{\sqrt{x}}\end{cases}}\)
Sau do the vo giai nhu binh thuong :D
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
5(x - 2) - 7(x - 3) = 4
<=> 5x - 10 - 7x + 21 = 4
<=> -2x + 11 = 4
=> -2x = 15
=> x = 15/-2
Ta có: \(5\left(x-2\right)-7\left(x-3\right)=4\)
\(\Leftrightarrow5x-10-7x+21=4\)
\(\Leftrightarrow-2x+11=4\)
\(\Leftrightarrow-2x=-7\)
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
b) Ta có: \(-2\left(x-5\right)-23=4-\left(x-2\right)\)
\(\Leftrightarrow-2x+10-23=4-x+2\)
\(\Leftrightarrow-2x-13=6-x\)
\(\Leftrightarrow-2x+x=6+13\)
\(\Leftrightarrow-x=19\)
\(\Leftrightarrow x=-19\)
Vậy \(x=-19\)
c) \(3\left(x-\frac{1}{2}\right)+\frac{5}{3}=\frac{6}{4}-\frac{1}{3}\left(x-2\right)\)
\(\Leftrightarrow3x-\frac{3}{2}+\frac{5}{3}=\frac{6}{4}-\frac{1}{3}x+\frac{2}{3}\)
\(\Leftrightarrow3x+\frac{1}{6}=-\frac{1}{3}x+\frac{13}{6}\)
\(\Leftrightarrow3x+\frac{1}{3}x=\frac{13}{6}-\frac{1}{6}\)
\(\Leftrightarrow\frac{10}{3}x=2\)
\(\Leftrightarrow x=\frac{3}{5}\)
Vậy \(x=\frac{3}{5}\)
- x - 2/3 = - 6/7
=> - x = - 6/7 + 2/3
=> - x = - 4/21
=> x = 4/21
Vậy x = 4/21
-x - 2/3 = - 6/7
- x = - 6/7 + 2/3
- x = - 4/21
x = 4/21
Vậy x=4/21
#Emma Nguyen