a: x-56:4=16

nên x-14=16

hay x=30

b: \(101+\left(36-4x\right)=105\)

\(\Leftrightarrow36-4x=4\)

hay x=8

Vì a: x-56:4=16

    =>x-14=16

   Hay:x=30

Vì b: 

   Hay:x=8

         Vậy x={30:8}

 Nếu sai thì thui nhá,đừng chửi mình!

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

2: \(x^2-2xy+y^2-2x+2y\)

\(=\left(x-y\right)^2-2\left(x-y\right)\)

=(x-y)(x-y-2)

3: \(3x^2-2x-5\)

\(=3x^2-5x+3x-5\)

=x(3x-5)+(3x-5)

=(3x-5)(x+1)

4: \(16-x^2+4xy-4y^2\)

\(=16-\left(x^2-4xy+4y^2\right)\)

\(=4^2-\left(x-2y\right)^2\)

=(4-x+2y)(4+x-2y)

5: \(x^2-2x+1-y^2\)

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

=(x-1-y)(x-1+y)

6: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

7: \(\left(x^2+6x+8\right)\left(x^2+14x+48\right)-9\)

=(x+2)(x+4)(x+6)(x+8)-9

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)-9\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+384-9\)

\(=\left(x^2+10x\right)^2+15\left(x^2+10x\right)+25\left(x^2+10x\right)+375\)

\(=\left(x^2+10x+25\right)\left(x^2+10x+15\right)=\left(x+5\right)^2\cdot\left(x^2+10x+15\right)\)

8: \(\left(x^2-8x+15\right)\left(x^2-16x+60\right)-24x^2\)

\(=\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+143x^2-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+119x^2\)

\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)

\(=\left(x^2-7x+30\right)\left(x-2\right)\left(x-15\right)\)

*vn:vô nghiệm.

a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

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

=>\(x^4-5x^2-2x+3=0\)

=>\(x^4+x^3-x^2-x^3-x^2+x-3x^2-3x+3=0\)

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

TH1: \(x^2+x-1=0\)

=>\(x^2+x+\frac14=\frac54\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(x+\frac12=\pm\frac{\sqrt5}{2}\)

=>\(x=-\frac12\pm\frac{\sqrt5}{2}\)

TH2: \(x^2-x-3=0\)

=>\(x^2-x+\frac14-\frac{13}{4}=0\)

=>\(\left(x-\frac12\right)^2=\frac{13}{4}\)

=>\(x-\frac12=\pm\frac{\sqrt{13}}{2}\)

=>\(x=\frac12\pm\frac{\sqrt{13}}{2}\)

c: \(3x^3+3x^2+3x=-1\)

=>\(x^3+3x^2+3x+1=-2x^3\)

=>\(\left(x+1\right)^3=\left(x\cdot\sqrt[3]{-2}\right)^3\)

=>\(x+1=x\cdot\sqrt[3]{-2}\)

=>\(x\left(1-\sqrt[3]{-2}\right)=-1\)

=>\(x=\frac{-1}{1-\sqrt[3]{-2}}\)

d: \(8x^3-12x^2+6x-5=0\)

=>\(8x^3-12x^2+6x-1-4=0\)

=>\(\left(2x-1\right)^3=4\)

=>\(2x-1=\sqrt[3]{4}\)

=>\(2x=1+\sqrt[3]{4}\)

=>\(x=\frac12+\frac12\cdot\sqrt[3]{4}\)

1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)

\(\Leftrightarrow2x-6+5⋮x-3\)

\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)

Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)

\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)

\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)

Vậy:...................

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)