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c) \(\left(3x+5\right)^2-2\left(2x+3\right)\left(3x+5\right)+\left(2x+3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left[\left(3x+5\right)-\left(2x+3\right)\right]^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(3x+5-2x-3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{-2;-1\right\}\)
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
tìm x
a.(x−2)\(^3\)−(x−3)(x\(^2\)+3x+9)+6(x+1)\(^2\)
b.(x+2)(x\(^2\)−2x+4x\(^2\)−2x+4)-x(x\(^2\)+2)=15
Câu a thiếu kết quả để tìm
Câu b)
(x + 2)(x2 - 2x + 4x2 - 2x + 4) - x(x2 + 2) = 15
=> (x + 2)(x2 - 4x + 4x2 + 4) - x3 + 2x = 15
=> (x + 2)(5x2 - 4x + 4) - x3 + 2x = 15
=> x(5x2 - 4x + 4) + 2(5x2 - 4x + 4) - x3 + 2x = 15
=> 5x3 - 4x2 + 4x + 10x2 - 8x + 8 - x3 + 2x = 15
=> (5x3 - x3) + (-4x2 + 10x2) + (4x - 8x + 2x) + 8 = 15
=> 4x3 + 6x2 - 2x + 8 = 15
=> 2(2x3 + 3x2 - x + 4) = 15
=> (2x3 + 3x2 - x + 4) = 15/2 => vô nghiệm