(2x/7)+(3-2x/7)
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a, \(\left(2x-7\right)^2+\left(2x+7\right)^2-2\left(2x+7\right)\left(2x-7\right)=\left(2x-7-2x-7\right)^2=\left(-14\right)^2=196\)
b, \(\left(5x-3\right)^2-\left(5x+3\right)^2-15\left(2x-1\right)\)
\(=\left(5x-3-5x-3\right)\left(5x-3+5x+3\right)-15\left(2x-1\right)\)
\(=-6.10x-15\left(2x-1\right)\)
\(=-60x-15\left(2x-1\right)=-15\left(4x+2x-1\right)=-15\left(6x-1\right)=-90x+15\)
\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow\frac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x+7\right)}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Rightarrow\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)=1-6x^2+9x-9\)
\(\Leftrightarrow4x^2+20x+21-4x^2-12x-5=1-6x^2+9x-9\)
\(\Leftrightarrow8x-16=1-6x^2+9x-9\)
\(\Leftrightarrow8x-16-1+6x^2-9x+9=0\)
\(\Leftrightarrow6x^2-x-8=0\)
Tự làm nốt nha
Trl
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nhé bạn ~
Đặt \(\sqrt{2x^3+7}=a\)
=>6ax=3a^2+1+2x-4a
=>a=2x+1 hoặc a=1/3
=>2x^3+7=(2x+1)^2 hoặc 2x^3+7=1/3
=>\(x\in\left\{1;\dfrac{1-\sqrt{13}}{2};\sqrt[3]{-\dfrac{31}{9}}\right\}\)
\(7^{2x}+7^{2x+3}=344\)
\(7^{2x}+7^{2x}.7^3=344\)
\(7^{2x}+7^{2x}.343=344\)
\(7^{2x}.1+7^{2x}.343=344\)
\(7^{2x}.\left(1+343\right)=344\)
\(7^{2x}.344=344\)
\(7^{2x}=344:344\)
\(7^{2x}=1\)
\(\Rightarrow7^{2x}=7^0\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
a) 3x - 4 (2x-7) = 3x - 3
3x - 8x + 28 = 3x - 3
=> 3x - 8x - 3x = -3 -28
-8x = -31
x = 31/8
b) 3(2x-7) - (4-2x) = 2x -1
6x - 7 - 4 + 2x = 2x -1
=> 6x + 2x - 2x = -1 + 7+4
6x = 10
x = 5/3
c) (x-2).(2x+12)=0
=> x -2 = 0 => x = 2
2x + 12 = 0 => 2x = -12 => x = -6
KL: x = 2 hoặc x = -6
=6x/7-4/49x^2