a. 3^x=1-x^2

x=0 la nghiem

x>=1;  VT>=3 VP<=0 vo nghiem 

b. (de bai thieu n khac 0 vi neu n=0 dung voi moi x)

3x-14=1=> x=5

c.(5^2x5^^x+1)=5^4

5^{^x+1}=5^2=> x=1

h(x)= x^4+4x^2-x^2-4x

      = (x^4-x^2) + (4x^2-4x)

      = x^2(x^2-1) + 4(x^2-1)

      = (x^2+4)(x^2-1)

Do đó ta có: h(x)=0 hay (x^2+4)(x^2-1)=0

                           Suy ra           x^2-1=0 (vì x^2+4 >0)

                                                x^2   =1

                                             =>x=1 hay x= -1.

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

\(\Rightarrow3x^2-3^2-3x^2+15x=1\)

\(\Rightarrow3x^2-9-3x^2+15x=1\)

\(\Rightarrow-9+15x=1\)

\(\Rightarrow15x=-8\)

\(\Rightarrow x=\frac{-8}{15}\)

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

\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)

\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)

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

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

\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)

\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)

\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)

\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)

\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)

-Bạn kiểm tra lại đề.

Simplifying 5(2x + 1) = 3(x + -2) Reorder the terms: 5(1 + 2x) = 3(x + -2) (1 * 5 + 2x * 5) = 3(x + -2) (5 + 10x) = 3(x + -2) Reorder the terms: 5 + 10x = 3(-2 + x) 5 + 10x = (-2 * 3 + x * 3) 5 + 10x = (-6 + 3x) Solving 5 + 10x = -6 + 3x Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-3x' to each side of the equation. 5 + 10x + -3x = -6 + 3x + -3x Combine terms: 10x + -3x = 7x 5 + 7x = -6 + 3x + -3x Combine terms: 3x + -3x = 0 5 + 7x = -6 + 0 5 + 7x = -6 Add '-5' to each side of the equation. 5 + -5 + 7x = -6 + -5 Combine terms: 5 + -5 = 0 0 + 7x = -6 + -5 7x = -6 + -5 Combine terms: -6 + -5 = -11 7x = -11 Divide each side by '7'. x = -1.571428571 Simplifying x = -1.571428571

ko hieu ban oi

đề bài là j vậy bạn

a, (3x-1)(x²+2)=(3x-1)(7x-10)

<=>(3x-1)(x²+2)-(3x-1)(7x-10)=0

<=>(3x-1)(x²+2-7x+10)=0

<=>(3x-1)(x²-7x+12)=0

<=>(3x-1)(x²-3x-4x+12)=0

<=>(3x-1)(x-3)(x-4)=0

<=>\(\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy ft có tập nghiệm S=\(\left\{\dfrac{1}{3},3,4\right\}\)

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (ĐKXĐ:t\(\ne2;t\ne-3\))

<=>\(\dfrac{\left(t+3\right)^2+\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{t^2-2t+3t-6}\)

<=>\(\dfrac{t^2+6t+9+t^2-4t+4}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

=>2t²+2t+13=5t+15

<=>2t²+2t-5t+13-15=0

<=>2t²-3t-2=0

<=>2t²-4t+t-2=0

<=>(t-2)(2t+1)=0

<=>\(\left[{}\begin{matrix}t-2=0\\2t+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-1}{2}\left(tmđkxđ\right)\end{matrix}\right.\)

Vậy ft có nghiệm duy nhất x=\(\dfrac{-1}{2}\)

Giải:

a) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

Chia cả hai vế cho 3x-1, ta được:

\(x^2+2=7x-10\)

\(\Leftrightarrow x^2-7x+10+2=0\)

\(\Leftrightarrow x^2-7x+12=0\)

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

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy ...

b) \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (1)

ĐKXĐ: \(t\ne2;t\ne-3\)

\(\left(1\right)\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

\(\Rightarrow\left(t+3\right)^2+\left(t-2\right)^2=5t+15\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)

\(\Leftrightarrow2t^2+2t+13=5t+15\)

\(\Leftrightarrow2t^2+2t+13-5t-15=0\)

\(\Leftrightarrow2t^2-3t-2=0\)

\(\Leftrightarrow2t^2-4t+t-2=0\)

\(\Leftrightarrow2t\left(t-2\right)+\left(t-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)

Vậy ...

a,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-7x+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1\\\left(x-4\right)\left(x-3\right)=0\end{matrix}\right.\)

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

Vậy...

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)

\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Leftrightarrow\dfrac{\left(t+3\right)^2}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)

\(\Leftrightarrow t^2+t^2+6t-4t-5t=15-9-4\)

\(\Leftrightarrow2t^2-3t=2\)

\(\Leftrightarrow\left(2t+1\right)\left(t+2\right)=0\)

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

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

Vậy...