a =>5x(x²-6x+9)-5(x³-3x²+3x-1)+15(x²-4)=5

=>5x³-30x²+45x-5x³+15x²+15x+5+15²-50=5
=>60x-55=5
=>x=1

c) x² ( x² +1 ) - x² -1 =0

x² (x² +1) -(x² +1) =0

(x² +1)(x² -1) =0

*) x² = -1 --> x không có giá trị thỏa mãn

*) x² = 1 --> x = 1

Vậy x= 1

a/ \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60-5=0\)

\(\Leftrightarrow30x-60=0\)

\(\Leftrightarrow30x=60\)

\(\Leftrightarrow x=2\)

vậy x=2

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

\(\Leftrightarrow3x-4x^2+6-8x=x^2+4x+4\)

\(\Leftrightarrow x^2+4x^2+4x+18x-3x+4-6=0\)

\(\Leftrightarrow5x^2+9x-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-2\end{matrix}\right.\)

vậy \(x=\dfrac{1}{5}\) hoặc \(x=-2\)

c/ \(x^2\left(x^2+1\right)-x^2-1=0\)

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

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

vì x²+1 >0 nên x² - 1 = 0 \(\Rightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

vậy \(x=1\) hoặc \(x=-1\)

1) 3x - 6 = 5x + 2

=> 3x - 5x = 2 + 6

=> -2x = 8 

=> x = -4

2) 15  - x = 4x - 5

=> 15 + 5 = 4x + x 

=> 20 = 5x

=> x = 4

3) x - 15 = 6 + 4x

=> x - 4x = 6 + 15

=> -3x = 21

=> x = -7

4) -12 + x = 5x - 20 

=> x - 5x = -20 + 12

=> -4x = -8

=> x = 2

5) 7x - 4 = 20 + 3x

=> 7x - 3x = 20 + 4

=> 4x = 24

=> x = 6

1) 3x-  6 = 5x + 2

5x - 3x = -6 - 2

2x = -8 => x = -4

Tương tự như trên 

Bài 1:

a: \(M=x^2+4x+4+5=\left(x+2\right)^2+5>=5\)

Dấu '=' xảy ra khi x=-2

b: \(N=x^2-20x+101=x^2-20x+100+1=\left(x-10\right)^2+1>=1\)

Dấu '=' xảy ra khi x=10

a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Sao lai co 3t(t-5) ,cho do thua

a) /x+\(\frac{4}{15}\)/ - / -3,75/ = -2,15

=> \(\orbr{\begin{cases}x+\frac{4}{15}+3,75=-2,15\\x+\frac{4}{15}+3,75=2,15\end{cases}}\)

=> ....v.....v giải ra ( từng th )

bài khác tương tự

ai biet tra loi cho mik voi mik dag rat can gap

\(\left(4x+1\right)\left(1-4x+16x^2\right)-16x\left(4x^2-5\right)=17\)

\(\Leftrightarrow4x-16x^2+64x^2+1-4x+16x^2-64x^2+80x-17=0\)

\(\Leftrightarrow\left(-16x^2+16x^2\right)+\left(64x^2-64x^2\right)+\left(4x-4x\right)+80x+1-17=0\)

\(\Leftrightarrow80x=16\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

Thứ nhất: Làm chi tiết ra k dc ạ?

Thứ 2: Kết quả sai. Xem lại.

1) 4x(x-5)-(x-1)(4x-3)=5

<=>4x²-20x-4x²+3x+4x-3=5

<=>-13x=8

<=>x=-8/13

Thôi mỏi tay quá tìm x luôn nha

2) x=1.875

3) x=17/7

cho mình hỏi bạn làm kiểu gì vậy