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anh ơi, vậy là sai đề hả anh, chứ đề kêu chứng minh phương trình vô nghiệm mà em thấy anh ghi x=2
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
Câu a:
\(2x\left(8x-1\right)^2\left(4x-1\right)=9\)
\(\Leftrightarrow\left(64x^2-16x+1\right)\left(64x^2-16x\right)=72\)
Đặt 64x2 - 16x = t \(\left(t\ge-1\right)\)
\(\Rightarrow t\left(t+1\right)=72\)
\(\Leftrightarrow\left(t+9\right)\left(t-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-9\left(loai\right)\\t=8\left(nhan\right)\end{matrix}\right.\)
\(\Rightarrow64x^2-16x=8\)
\(\Leftrightarrow8\left(2x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Câu b:
\(\Leftrightarrow\left(x+1\right)^2\left(2x+1\right)\left(2x+3\right)=18\)
\(\Leftrightarrow\left(4x^2+8x+4\right)\left(4x^2+8x+3\right)=72\)
Đặt 4x2 + 8x + 4 = m \(\left(m\ge0\right)\)
\(\Rightarrow m\left(m-1\right)=72\)
\(\Leftrightarrow\left(m-9\right)\left(m+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=9\left(nhan\right)\\m=-8\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow4\left(x+1\right)^2=9\)
\(\Leftrightarrow x+1=\pm\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow2x\cdot\left(64x^2-16x+1\right)\cdot\left(4x-1\right)=9\)
\(\Rightarrow\left(64x^2-16x+1\right)\cdot\left(8x^2-2x\right)=9\)
Nhân cả hai vế của phương trình với 8 ta được:
\(\left(64x^2-16x+1\right)\cdot\left(64x^2-16x\right)=72\)
Đặt \(a=64x^2-16x\left(a\ge1\right)\) (1)
\(\Rightarrow\left(a+1\right)\cdot a=72\)
\(\Rightarrow a^2+a-72=0\)
\(\Rightarrow\left(a-8\right)\cdot\left(a+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=8\left(tmđk\right)\\a=-9\left(loại\right)\end{matrix}\right.\)
Thay vào (1) ta đc:
\(64x^2-16x=8\Rightarrow64x^2-16x-8=0\)
\(\Rightarrow\left(2x-1\right)\left(4x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
a) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)
⇒ \(8x + 3x + 3 > 5x - 2x + 6\)
⇒ \(11x+3>3x+6\)
⇒ \(11x - 3x > 6 -3\)
⇒ \(8x > 3\)
⇒ \(8x.\dfrac{1}{8}>3.\dfrac{1}{8}\)
⇒ \(x>\dfrac{3}{8}\)
S = \(\left\{x\backslash x>\dfrac{3}{8}\right\}\)
b) \(2x(6x-1) > (3x -2)(4x+3)\)
⇒ \(12x^2 - 2x > 12x^2 +9x -8x -6\)
⇒ \(12x^2 - 2x > 12x^2 + x - 6\)
⇒ \(-2x-x>12x^2 -6-12x^2\)
⇒ \(- 3x > -6 \)
⇒ \(x > 2\)
S = {x / x > 2}
Sửa đề: 8x-1
=>2(8x^2-x)(8x^2-x+2)-126=0
=>2[(8x^2-x)^2+2(8x^2-x)]-126=0
=>(8x^2-x)^2+2(8x^2-x)-63=0
=>(8x^2-x+9)(8x^2-x-7)=0
=>8x^2-x-7=0
=>x=1 hoặc x=-7/8
\(\Leftrightarrow8x\left(8x-1\right)^2\left(8x-2\right)=72.\)(nhân cả 2 vế vs 8)
Đặt \(a=8x-1.\)ta có pt
\(\left(a-1\right)a^2\left(a+1\right)=72\)
\(\Leftrightarrow a^4-a^2-72=0\)
\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0.\)
\(\Rightarrow\left(a-3\right)\left(a+3\right)=0\)(do \(a^2+8\ne0.\))
\(\Rightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}8x-1=3\\8x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0.5\\x=-0.25\end{cases}}\)
vậy, \(S=\left\{0.5;-0.25\right\}.\)
xong rồi đó bn
ko có dấu cộng hay dấu trừ j ak