Giải phương trình (4x+3)^3+(5-7x)^3+(3x-8)^3
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<=>(4x-3)3+(5-7x)3+(3x-8)3=-3(3x-8)(4x+3)(7x-5)
=>-3(3x-8)(4x+3)(7x-5)=0
Th1:-3(3x-8)=0
=>3x-8=0
=>3x=8
=>x=\(\frac{8}{3}\)
Th2:4x+3=0
=>4x=-3
=>x=\(-\frac{3}{4}\)
Th3:7x-5=0
=>7x=5
=x=\(\frac{5}{7}\)
Đặt 4x+3=a; 3x-8=b
Theo đề, ta có phương trình:
\(a^3+b^3-\left(a+b\right)^3=0\)
=>3ab(a+b)=0
=>(4x+3)(3x-8)(7x-5)=0
hay \(x\in\left\{-\dfrac{3}{4};\dfrac{8}{3};\dfrac{5}{7}\right\}\)
Đặt \(\hept{\begin{cases}x^2+3x-4=a\\3x^2+7x+4=b\end{cases}\Rightarrow4x^2+10x=a+b}\)
\(\left(x^2+3x-4\right)^3+\left(3x^2+7x+4\right)^3=\left(4x^2+10x\right)^3\)
\(\Rightarrow a^3+b^3=\left(a+b\right)^3\)
\(\Rightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow3ab\left(a+b\right)=0\)
Nếu \(a=0\Rightarrow x^2+3x-4=0\Rightarrow x\left(x+4\right)-\left(x+4\right)=0\Rightarrow\left(x+4\right)\left(x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)
Nếu \(b=0\Rightarrow3x^2+7x+4=0\Rightarrow3x\left(x+1\right)+4\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(3x+4\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{4}{3}\end{cases}}\)
Nếu \(a+b=0\Rightarrow4x^2+10x=0\Rightarrow2x\left(2x+5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}\)
a: =>(x-3)(x+1)=0
=>x=3 hoặc x=-1
b: =>x(x-3)=0
=>x=0 hoặc x=3
c: =>(x-5)(x+1)=0
=>x=5 hoặc x=-1
d: =>5x^2+7x-5x-7=0
=>(5x+7)(x-1)=0
=>x=1 hoặc x=-7/5
e: =>x^2-4=0
=>x=2 hoặc x=-4
h: =>x^2-4x+4-3=0
=>(x-2)^2=3
=>\(x=2\pm\sqrt{3}\)
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
Các bước giải:
b1: cho biểu thức =0
(4x+3)3+(5-7x)3+(3x-8)3=0
b2: Dùng hằng đẳng thức A3-B3 và A3+B3
b3: Chuyển đổi
Nhớ like nha