\(5x\left(6-x\right)+\left(5x-3\right)x=27\)

\(30x-5x^2+5x^2-3x=27\)

\(27x=27\)

\(x=1\)

Vậy \(x=1\)

\(3\left(2x-1\right)-5\left(x-3\right)+6.\left(3x-4\right)=83\)

\(6x-3-5x+15+18x-24=83\)

\(19x-12=83\)

\(19x=95\)

\(x=\frac{95}{19}\)

\(x=5\)

Vậy \(x=5\)

\(5x.\left(6-x\right)+\left(5x-3\right)x=27\)

\(\Rightarrow30x-5x^2+5x^2-3x=27\)

\(\Rightarrow\left(30x-3x\right)-\left(5x^2-5x^2\right)=27\)

\(\Rightarrow27x=27\)

\(\Rightarrow x=\)................................................(XIn lỗi em không biết chia :D)

\(a,5x\left(6-x\right)+\left(5x-3\right)x=27\)

\(\Leftrightarrow30x-5x^2+5x^2-3x=27\)

\(\Leftrightarrow27x=27\)

\(\Leftrightarrow x=1\)

\(b,3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)=83\)

\(\Leftrightarrow6x-3-5x+15+18x-24=83\)

\(\Leftrightarrow19x-12=83\)

\(\Leftrightarrow19x=95\)

\(\Leftrightarrow x=5\)

\(5x\left(6-x\right)+x\left(5x-3\right)=5.\)

\(30x-5x^2+5x^2-3x=5\)

\(27x=5\Rightarrow x=\frac{5}{27}\)

(5x + 3⁴).6⁸ = 6⁹.3⁴

5x + 3⁴ = 6⁹.3⁴ : 6⁸

5x + 81 = 6.81

5x = 6.81 - 81

5x = 81.(6 - 1)

5x = 81.5

x = 81.5 : 5

x = 81

Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

\(\left(x^2-5x\right)+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

a) Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)

\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b,\(< =>25x^2+10x+1-25x^2+9-30=0\)

\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)

c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)

\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)

\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)

\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)

a: Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

a) \(M(x) = A(x) + B(x) \\= 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4} \\=(4x^4-4x^4)+(-7x^3+7x^3)+(6x^2-5x^2)+(-5x+5x)+(-6+4)\\= {x^2} - 2.\)

b) \(A(x) = B(x) + C(x) \Rightarrow C(x) = A(x) - B(x)\)

\(\begin{array}{l}C(x) = A(x) - B(x)\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - ( - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4})\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 + 5{x^2} - 7{x^3} - 5x - 4 + 4{x^4}\\ =(4x^4+4x^4)+(-7x^3-7x^3)+(6x^2+5x^2)+(-5x-5x)+(-6-4)\\= 8{x^4} - 14{x^3} + 11{x^2} - 10x - 10\end{array}\)