2^x+1 : 4=32
3^x-2:3=243
256:4^x+1=4^2
4^2x-1:4=4^4
5^x-1:5=5^3
3^2x+1:3=3^4
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1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
a, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{3}{2}\Rightarrow5x+6x=-15\Leftrightarrow x=-\dfrac{15}{11}\)
b, TH1 : \(\dfrac{2}{3}x-\dfrac{4}{7}=0\Leftrightarrow x=\dfrac{6}{7}\);TH2 : \(\dfrac{1}{2}-\dfrac{3}{7x}=0\Rightarrow7x-6=0\Leftrightarrow x=\dfrac{6}{7}\)
c, TH1 : \(\dfrac{4}{5}-2x=0\Leftrightarrow x=\dfrac{4}{5}:2=\dfrac{2}{5}\)
TH2 : \(\dfrac{1}{3}+\dfrac{3}{5x}=0\Rightarrow5x+9=0\Leftrightarrow x=-\dfrac{9}{5}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
1,\(=4x\left(x-\dfrac{3}{2}\right)\)
2,\(=-7y^3\left[2x^2y\left(2y+x\right)+3\right]\)
3, = 4x(a-b)-6xy(a-b)
=2x(a-b)(2-3y)
4,
=3(2x+1)-(2x-5)(2x+1)
=(3-2x+5)(2x+1)
=(8-2x)(2x+1)
=2(4-x)(2x+1)
1 ) \(\left(2x-5\right)+17=6\)
\(\Leftrightarrow2x-5=-11\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3.\)
2 ) \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow2+6x=-4\)
\(\Leftrightarrow6x=-6\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1.\)
3 ) \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow3x+9=-18\)
\(\Leftrightarrow3x=-27\)
\(\Leftrightarrow x=-9\)
Vậy \(x=-9.\)
4 ) \(24:\left(3x-2\right)=-3\)
\(\Leftrightarrow\left(3x-2\right)=-8\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2.\)
5 ) \(-45:5.\left(-3-2x\right)=3\)
\(\Leftrightarrow-9\left(-3-2x\right)=3\)
\(\Leftrightarrow18x+27=3\)
\(\Leftrightarrow18x=-24\)
\(\Leftrightarrow x=-\frac{4}{3}\)
Vậy \(x=-\frac{4}{3}.\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
\(2^x:4=32\\ \Rightarrow2^x=128\\ \Rightarrow2^x=2^7\\ \Rightarrow x=7\)
\(3^{x-2}:3=243\\ \Rightarrow3^{x-2}=729\\ \Rightarrow3^{x-2}=3^6\\ \Rightarrow x-2=6\\ \Rightarrow x=8\)
\(256:4^{x+1}=4^2\\ \Rightarrow4^{x+1}=4^2\\ \Rightarrow x+1=2\\ \Rightarrow x=1\)
\(4^{2x-1}:4=4^4\\ \Rightarrow4^{2x-1}=4^5\\ \Rightarrow2x-1=5\\ \Rightarrow x=3\)
\(5^{x-1}:5=5^3\\ \Rightarrow5^{x-1}=5^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
\(3^{2x+1}:3=3^4\\ \Rightarrow3^{2x+1}=3^5\\ \Rightarrow2x+1=5\\ \Rightarrow x=3\)