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\(\left(4x-1\right)\left(4x-1\right)^2-\left(4x-3\right)\left(16x^2+3\right)\)
\(=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)
\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2-9\)
=-10
3/4x-7/3=1/4x+1/6
=>3/4x=1/4x+1/6+7/3
=>3/4x=1/4x+5/2
=>3/4x - 1/4x =5/2
=>1/2x=5/2
=>.......................................
4x + 3 + 4x = 1040
4x . 43 + 4x . 1 = 1040
4x ( 43 + 1 ) = 1040
4x . 65 = 1040
4x = 1040 : 65 = 16 = 42
Vậy x = 2
\(4^{x+3}+4^x=1040\)
\(\Leftrightarrow4^x\left(1+4^3\right)=1040\)
\(\Leftrightarrow4^x.65=1040\)
\(\Leftrightarrow4^x=16\)
\(\Leftrightarrow x=2\)
1)8(4x-3)-3(2-3x)=13-40x
(32x-24)-(6-9x)=13-40x
32x-24-6+9x=13-40x
41x-30=13-40x
41x+40x=13+30
81x=43
x=43/81
Vậy x=43/81
2)10x-5(1-4x)=5x-11
10x-(5-20x)=5x-11
10x-5+20x=5x-11
30x-5=5x-11
30x-5x=5-11
25x=-6
x=-6/25
Vậy x=-6/25
320-4x+43=352
4x+64=320-352
4x+64=-32
4x=-32-64
4x=-96
x=-96:4
x=-24
4x - 3 = 4x - ( 8 - 5 ) = 4x - 8 + 5 = 4x - 4.2 + 5 = 4.( x - 2 ) + 5
k cho mk nha bạn
Sửa đề : \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)-\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6-\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=\frac{7}{4}x+2x\)
\(-x-\frac{1}{4}=2x\)
\(-x-\frac{1}{4}-2x=0\)
\(-3x-\frac{1}{4}=0\)
\(-3x=\frac{1}{4}\Leftrightarrow x=-\frac{1}{12}\)
\(2\left(\frac{3}{4}x-1\right)-3y\left(1-\frac{3}{4}x\right)=0\)
\(\Leftrightarrow\)\(2\left(\frac{3}{4}x-1\right)+3y\left(\frac{3}{4}x-1\right)=0\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-1\right)\left(2+3y\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{3}{4}x-1=0\\2+3y=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{4}{3}\\y=-\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Leftrightarrow\frac{-3}{2}x=\frac{-23}{4}\)
\(\Leftrightarrow x=\frac{23}{6}\)
4x3 = 4x
4x3 - 4x = 0
4x(x2 - 1) = 0
\(\Rightarrow\orbr{\begin{cases}4x=0\\x^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1;1\end{cases}}\)
Vậy x = { - 1;0; 1 }
thak you ạ