\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)

\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)

\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)

\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)

\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)

\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)

\(\Leftrightarrow4x^2+6x-51=0\)

\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)

chuyển vế sang r phân tích thành nhân tử, có thể dùng máy tính bỏ túi nhé bạn

câu 1: 9\(x^2\) + 12\(x\) + 5  =11

           (3\(x\))² + 2.3.\(x\) .2 + 2² + 1 = 11

           (3\(x\) + 2)²      =  11 - 1

             (3\(x\) + 2)²    = 10

               \(\left[{}\begin{matrix}3x+2=\sqrt{10}\\3x+2=-\sqrt{10}\end{matrix}\right.\)

                \(\left[{}\begin{matrix}3x=\sqrt{10}-2\\3x=-\sqrt{10}-2\end{matrix}\right.\)

                  \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{3}\\x=\dfrac{-\sqrt{10}-2}{3}\end{matrix}\right.\)

                 Vậy S = {\(\dfrac{-\sqrt{10}-2}{3}\); \(\dfrac{\sqrt{10}-2}{3}\)} 

  Câu 2: 6\(x^2\) + 16\(x\) + 12 = 2\(x^2\)

              6\(x^2\) + 16\(x\) + 12 - 2\(x^2\) = 0

              4\(x^2\) + 16\(x\) + 12 = 0

              (2\(x\))² + 2.2.\(x\).4 + 16 - 4 = 0

               (2\(x\) + 4)²   = 4

               \(\left[{}\begin{matrix}2x+4=2\\2x+4=-2\end{matrix}\right.\) 

                \(\left[{}\begin{matrix}2x=-2\\2x=-6\end{matrix}\right.\)

                 \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

              S = { -3; -1}

3, 16\(x^2\) + 22\(x\) + 11 = 6\(x\) + 5

    16\(x^2\) + 22\(x\) - 6\(x\)  + 11 - 5 = 0

     16\(x^2\) + 16\(x\) + 6 = 0

      (4\(x\))² + 2.4.\(x\) . 2 + 2² + 2 = 0

       (4\(x\) + 2)² + 2 = 0 (1) 

Vì (4\(x\)+ 2)² ≥ 0 ∀ ⇒ (4\(x\) + 2)² + 2 > 0 ∀ \(x\) vậy (1) Vô nghiệm

             S = \(\varnothing\)

Câu 4. 12\(x^2\) + 20\(x\) + 10 = 3\(x^2\) - 4\(x\) 

            12\(x^2\) + 20\(x\) + 10 - 3\(x^2\) + 4\(x\) = 0

            9\(x^2\) + 24\(x\) + 10 = 0

           (3\(x\))² + 2.3.\(x\).4 + 16 - 6 = 0

          (3\(x\) + 4)² = 6

            \(\left[{}\begin{matrix}3x+4=\sqrt{6}\\3x+4=-\sqrt{6}\end{matrix}\right.\)

              \(\left[{}\begin{matrix}3x=-4+\sqrt{6}\\3x=-4-\sqrt{6}\end{matrix}\right.\)

              \(\left[{}\begin{matrix}x=\dfrac{\sqrt{6}-4}{3}\\x=-\dfrac{\sqrt{6}+4}{3}\end{matrix}\right.\)

                    S = {\(\dfrac{-\sqrt{6}-4}{3}\); \(\dfrac{\sqrt{6}-4}{3}\)}

a: =>6x-3x^2-5=4-3x^2-2

=>6x-5=2

=>6x=7

=>x=7/6

b: =>20x+5-12x^2-3x=6x^2-10x+3x-5

=>-12x^2+17x+5-6x^2+7x+5=0

=>-18x^2+24x+10=0

=>x=5/3 hoặc x=-1/3

Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

\(\left|2x-3\right|-4x-9=0\)

<=> \(\left|2x-3\right|=4x+9\)

<=> \(\orbr{\begin{cases}2x-3=4x+9\left(x\ge\frac{3}{2}\right)\\3-2x=4x+9\left(x< \frac{3}{2}\right)\end{cases}}\) <=> \(\orbr{\begin{cases}2x=-12\\6x=-6\end{cases}}\) <=> \(\orbr{\begin{cases}x=-6\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)

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

<=> \(\left|5-3x\right|=x^2+2x+1-x-x^2-2x-4\)

<=> \(\left|5-3x\right|=-x-3\)

<=> \(\orbr{\begin{cases}5-3x=-x-3\left(x\le\frac{5}{3}\right)\\5-3x=x+3\left(x>\frac{5}{3}\right)\end{cases}}\) <=> \(\orbr{\begin{cases}2x=8\\4x=2\end{cases}}\) <=> \(\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{2}\left(ktm\right)\end{cases}}\)

=> pt vô nghiệm

a: P(x)=2x^5-2x^5+4x^4-3x^4+5=x^4+5

Q(x)=-5x^4+2x^4-x^3+3x^2-10x+2

=-3x^4-x^3+3x^2-10x+2

b: P(x)+Q(x)

=x^4+5-3x^4-x^3+3x^2-10x+2

=-2x^4-x^3+3x^2-10x+7

Q(x)-P(x)

=-3x^4-x^3+3x^2-10x+2-x^4-5

=-4x^4-x^3+3x^2-10x-3

P(x)-Q(x)=-(Q(x)-P(x))

=4x^4+x^3-3x^2+10x+3

\(1,4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-4x^2-8-1+4x^2+3=0\\ \Leftrightarrow4x-6=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(2,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\\ \Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(2-3x\right)\left(5x-2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(x+11-5x+2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

d: \(\Leftrightarrow\left(4x+2\right)\left(x-1\right)-\dfrac{1}{2}x\left(x-1\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(-\dfrac{7}{2}x+2\right)=0\)

=>x=1 hoặc x=4/7

e: \(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)

=>10x-4=15-9x

=>19x=19

hay x=1

f: \(\Leftrightarrow\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)

=>2x-1+x-1=1

=>3x-2=1

hay x=1(loại)

g: =>1+3x-6=3-x

=>3x-5-3+x=0

=>4x-8=0

=>x=2(loại)