\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)

\(\Leftrightarrow7-6x=3x-5\)

\(\Leftrightarrow7+5=3x+6x\)

\(\Leftrightarrow12=9x\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

a) \(6x^3-6x=0\Leftrightarrow6x\left(x^2-1\right)=0\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)b) \(2x\left(3x+7\right)-6x^2=28\Leftrightarrow6x^2+14x-6x^2=28\Leftrightarrow14x=28\Leftrightarrow x=2\)

c) \(2\left(4x+4\right)-5\left(x-3\right)=0\Leftrightarrow8x+8-5x+15=0\Leftrightarrow3x=-23\Leftrightarrow x=-\dfrac{23}{3}\)

a: Ta có: \(6x^3-6x=0\)

\(\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

b: Ta có: \(2x\left(3x+7\right)-6x^2=28\)

\(\Leftrightarrow6x^2+14x-6x^2=28\)

\(\Leftrightarrow14x=28\)

hay x=2

c: Ta có: \(2\left(4x+4\right)-5\left(x-3\right)=0\)

\(\Leftrightarrow8x+8-5x+15=0\)

\(\Leftrightarrow3x=-23\)

hay \(x=-\dfrac{23}{3}\)

a) Ta có: \(x^3+6x-7\)

\(=x^3-x+7x-7\)

\(=x\left(x-1\right)\left(x+1\right)+7\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+7\right)\)

b) Ta có: \(4x^2+8x-5\)

\(=4x^2+10x-2x-5\)

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

\(=\left(2x+5\right)\left(2x-1\right)\)

c) Ta có: \(9x^2-4y^2+6x-4y\)

\(=9x^2+6x+1-\left(4y^2+4y+1\right)\)

\(=\left(3x+1\right)^2-\left(2y+1\right)^2\)

\(=\left(3x+1+2y+1\right)\left(3x+2y\right)\)

\(=\left(3x+2y\right)\left(3x+2y+2\right)\)

Đặt \(x^2+6x+7=t\)

Bài toán trở thành tìm m để phương trình: \(\left(t-2\right)\left(t+1\right)=m-1\) (1) có nghiệm \(t< 0\)

\(\left(1\right)\Leftrightarrow t^2-t-1=m\)

Xét hàm \(f\left(t\right)=t^2-t-1\)

\(f\left(0\right)=-1\) và hàm số nghịch biến khi \(t< 0\)

\(\Rightarrow f\left(t\right)>-1\) \(\forall t< 0\)

\(\Rightarrow\) phương trình \(f\left(t\right)=m\) có nghiệm \(t< 0\) khi và chỉ khi \(m>-1\)

Vậy với \(m>-1\) thì pt đã cho có nghiệm thỏa \(x^2+6x+7< 0\)

* Trả lời:

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

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

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

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

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

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

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ừ đúng dễ mà

Lời giải:
ĐKXĐ:......

Ta có:

\(x^2-6x+\sqrt{x^2-6x+7}=5\)

\(\Leftrightarrow x^2-6x+7+\sqrt{x^2-6x+7}=12\)

Đặt \(\sqrt{x^2-6x+7}=a(a\geq 0)\). Khi đó pt trở thành:

\(a^2+a=12\)

\(\Leftrightarrow a^2+a-12=0\)

\(\Leftrightarrow (a-3)(a+4)=0\Rightarrow a=3\) (do $a\geq 0$)

\(\Rightarrow x^2-6x+7=a^2=9\)

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

\(\Rightarrow x=3\pm \sqrt{11}\) (thỏa mãn)

Vậy........

\(ĐKXĐ:x^2-6x+7\ge0\)

ĐẶt: \(n=\sqrt{x^2-6x+7}\left(n\ge0\right)\)

\(\Leftrightarrow n^2-7=x^2-6x\)

Phương trình thành:

\(n^2-7+n=5\)

\(\Leftrightarrow n^2+n-12=0\)

\(\Leftrightarrow n^2+2\cdot n\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-12=0\)

\(\Leftrightarrow\left(n+\dfrac{1}{2}\right)^2-\dfrac{49}{4}=0\)

\(\Leftrightarrow\left(n+\dfrac{1}{2}\right)^2=\dfrac{49}{4}\)

\(\Rightarrow\left[{}\begin{matrix}n+\dfrac{1}{2}=\dfrac{7}{2}\\n+\dfrac{1}{2}=-\dfrac{7}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=3\\n=-4\end{matrix}\right.\)(loại n=-4)

Với n=3

\(\Rightarrow\sqrt{x^2-6x+7}=3\)

\(\Leftrightarrow x^2-6x+7=9\)

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

\(\Leftrightarrow x^2-2\cdot x\cdot3+9-9-2=0\)

\(\Leftrightarrow\left(x-3\right)^2=11\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{11}+3\\x=-\sqrt{11}+3\end{matrix}\right.\)

Giúp

\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{3}{7}\)

\(x=\dfrac{7}{9}\)

\(x:\dfrac{6}{8}=\dfrac{2}{5}\)

\(x=\dfrac{2}{5}\times\dfrac{6}{8}\)

\(x=\dfrac{6}{20}=\dfrac{3}{10}\)

\(x-\dfrac{2}{3}=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}+\dfrac{2}{3}\)

\(x=\dfrac{9}{6}=\dfrac{3}{2}\)

\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)

\(x=\dfrac{1}{7}+\dfrac{4}{35}+\dfrac{2}{5}\)

\(x=\dfrac{23}{35}\)

.

\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}:\dfrac{3}{7}\)

\(\Leftrightarrow x=\dfrac{7}{9}\)

\(x:\dfrac{6}{8}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}\times\dfrac{6}{8}\)

\(\Leftrightarrow x=\dfrac{3}{10}\)

\(x-\dfrac{2}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow x=\dfrac{5}{6}+\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)

\(\Leftrightarrow x=\dfrac{1}{7}+\dfrac{2}{5}+\dfrac{4}{35}\)

\(\Leftrightarrow x=\dfrac{23}{35}\)