\(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)
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15 tháng 4 2019
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
15 tháng 4 2019
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2
19 tháng 12 2016
1)
ĐKXĐ: x\(\ne\)3
ta có :
\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)
để biểu thức A có giá trị = 1
thì :\(\frac{x-3}{2}\)=1
=>x-3 =2
=>x=5(thoả mãn điều kiện xác định)
vậy để biểu thức A có giá trị = 1 thì x=5
HN
30 tháng 12 2016
1)
\(A=\frac{x^2-6x+9}{2x-6}\)
A xác định
\(\Leftrightarrow2x-6\ne0\)
\(\Leftrightarrow2x\ne6\)
\(\Leftrightarrow x\ne3\)
Để A = 1
\(\Leftrightarrow x^2-6x+9=2x-6\)
\(\Leftrightarrow x^2-6x-2x=-6-9\)
\(\Leftrightarrow x^2-8x=-15\)
\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)
5 tháng 5 2020
Câu 1:
PT <=> \(\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
<=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
<=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Mà \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
<=> x - 100 = 0
<=> x = 100
Câu 2
PT <=> \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
<=> \(\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Mà \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
<=> x + 100 = 0
<=> x = -100
Câu 3:
PT <=> \(\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)
<=> \(\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)
<=> \(\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)=0\)
mà \(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\ne0\)
<=> x+15 = 0
<=> x = -15
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 5 2020
1/
\(\Leftrightarrow\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=0\)
\(\Leftrightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
\(\Rightarrow x=100\)
2/
\(\frac{x+2}{98}+1+\frac{x+4}{96}+1-1-\frac{x+6}{94}-1-\frac{x+8}{92}=0\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Rightarrow x=-100\)
3/
\(\frac{x+2}{13}+1+\frac{2x+45}{15}-1-1-\frac{3x+8}{37}+1-\frac{4x+69}{9}=0\)
\(\Leftrightarrow\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}-\frac{3\left(x+15\right)}{37}-\frac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)=0\)
\(\Rightarrow x=-15\)
13 tháng 2 2020
ĐKXĐ:...
a) \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{3}\\\frac{x+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2y}{3}\\\frac{\frac{2y}{3}+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-12}{19}\\x=\frac{-8}{19}\end{matrix}\right.\)
Vậy...
b) \(\left\{{}\begin{matrix}0,75x-3,2y=10\\x\sqrt{3}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3,2y+10}{0,75}\\\frac{\left(3,2y+10\right)\sqrt{3}}{0,75}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{\frac{16\sqrt{3}}{5}y+10\sqrt{3}-\frac{3\sqrt{2}}{4}y}{0,75}=4\sqrt{3}\\x=\frac{3,2y+10}{0,75}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\left(\frac{16\sqrt{3}}{5}-\frac{3\sqrt{2}}{4}\right)+10\sqrt{3}=3\sqrt{3}\\x=\frac{3,2y+10}{0,75}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-140\sqrt{3}}{64\sqrt{3}-15\sqrt{2}}\\x=\frac{\frac{-448\sqrt{3}}{64\sqrt{3}-15\sqrt{2}}+10}{0,75}\end{matrix}\right.\)
Nghiệm đẹp lắm.
c) \(\left\{{}\begin{matrix}\frac{2x+3}{y-1}=\frac{4x+1}{2y+1}\\\frac{x+2}{y-1}=\frac{x-4}{y+2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(2y+1\right)-\left(y-1\right)\left(4x+1\right)=0\\\left(x+2\right)\left(y+2\right)-\left(y-1\right)\left(x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+5y+4=0\\3x+6y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2y\\-12y+5y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4}{7}\\x=\frac{-8}{7}\end{matrix}\right.\)
Vậy...
18 tháng 8 2016
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
\(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)
\(\Leftrightarrow\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)
\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)
\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}-\frac{x-73}{67}-\frac{x-73}{65}=0\)
\(\Leftrightarrow\left(x-73\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)
Mà \(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\ne0\)
\(x-73=0\Leftrightarrow x=73\)
\(\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)
\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)
\(\Rightarrow...\)