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1. <=>15 (2x+4)= 7(4x-2)
<=>30x+60= 28x-14
<=>2x= -74
<=>x= -37
2. <=> -5(12-7x)= -13(4-3x)
<=> -60+35x = -52+39x
<=> -4x= 8
<=> x= -2
3.<=> 88x-16 = 77x+55
<=> 11x= 71
<=> x=71/11
a)
\(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=5.\left(x+5\right)\)
\(\Leftrightarrow7x-21=25+5x\)
\(\Leftrightarrow7x-5x=25+21\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=23\)
b)
\(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=7.9\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x=8\)
Mẫy bài còn lại làm tương tự
\(c,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Leftrightarrow(x-1)(x+3)=(x-2)(x+2)\)
\(\Leftrightarrow x^2+2x-3=x^2-4\)
\(\Leftrightarrow x^2+2x-3-x^2=-4\)
\(\Leftrightarrow x^2-x^2+2x-3=-4\)
\(\Leftrightarrow2x-3=-4\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)
1) \(\frac{2x+3}{6}=\frac{7x-3}{-5}\)
\(-5\left(2x+3\right)=6\left(7x-3\right)\)
\(-10x-15=42x-18\)
\(52x=3\)
\(x=\frac{3}{52}\)
2) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
\(-5\left(12-7x\right)=-13\left(4-3x\right)\)
\(-60+35x=-52+39x\)
\(39x-35x=-60+52\)
\(4x=-8\)
\(x=-2\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}.\left(6x+1\right)\)
\(\Rightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}.6x+\frac{2}{3}\)
\(\Rightarrow\frac{55-24}{60}-x=4x+\frac{2}{3}\)
\(\Rightarrow\frac{31}{60}-x=4x+\frac{2}{3}\)
\(\Rightarrow\frac{31}{60}-\frac{2}{3}=4x+x=5x\)
\(\Rightarrow5x=-\frac{11}{60}\)
\(\Rightarrow x=\frac{-11}{300}\)
\(=\frac{2^{12}\times3^{10}+2^9\times3^9\times2^3\times3\times5}{-2^{12}\times3^{12}-2^{11}\times3^{11}}\)
\(=\frac{2^{12}\times3^{10}+2^{12}\times3^{10}\times5}{-2^{11}\times3^{11}\times\left(2\times3+1\right)}\)
\(=\frac{2^{12}\times3^{10}\times\left(1+5\right)}{-2^{11}\times3^{11}\times7}\)
\(=\frac{2^{13}\times3^{11}}{-2^{11}\times3^{11}\times7}\)
\(=\frac{-4}{7}\)
ta có 2TH
TH1 6x+10y+2z - 11 = 0
\(\Rightarrow\hept{\begin{cases}3x-2=0\\2y+1=0\\z-6=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{1}{2}\\z=6\end{cases}}}\)
TH2 6x+10y+2z - 11 \(\ne\)0
áp dụng t/c dãy tỉ số bằng nhau ta có
\(\frac{3x-2}{3}=\frac{2y+1}{4}=\frac{z-6}{5}=\frac{6x+10y+2z-11}{6+20+10}\)\(=\frac{6x+10y+2z-11}{36}\)
=> 36 = 11x + 3
=> x = 3
\(\Rightarrow\hept{\begin{cases}\frac{3x-2}{3}=\frac{7}{3}\\\frac{2y+1}{4}=\frac{7}{3}\\\frac{z-6}{5}=\frac{7}{3}\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=\frac{25}{6}\\z=\frac{53}{3}\end{cases}}\)
đề sai
nếu mà duy trì đề đó thì TA có:
11x - 2 = 11
11x = 11 + 2 = 13
x = 13 : 11
x = 1,181818182
_____________
7x + 5 = 8
7x = 8 - 5 = 3
x = 3 : 7
x = 0,428571428
mk nghĩ là nhân tích chéo.
(11x-2).8=11.(7x+5)
8.11x-2.8=11.7x+11.5
8.11x-11.7x=55+16
(8.11)-(11x-7x)=71
88-4x=71 4x=88-71 4x=17 x=17:4 x=17/4
\(\frac{11x-2}{7x-5}=\frac{11}{8}\)
\(\Leftrightarrow8\left(11x-2\right)=11\left(7x-5\right)\)
\(\Leftrightarrow88x-16=77x-55\)
\(\Leftrightarrow88x-77x=-55+16\)
\(\Leftrightarrow11x=-39\)
\(\Leftrightarrow x=\frac{-39}{11}\)
\(\frac{12x-2}{7x-8}=\frac{11}{8}\)
\(\Rightarrow8\left(12x-2\right)=11\left(7x-8\right)\)
\(\Rightarrow96x-16=77x-88\)
\(\Rightarrow96x-77x=-88+16\)
\(\Rightarrow19x=-72\)
\(\Rightarrow x=-\frac{72}{19}\)