a) Cho đa thức : x^2 – 4 = 0 => x^2 = 4

Vậy x = 2; x = – 2 là nghiệm của đa thức

b) Cho đa thức: x – x^2 = 0

– Phân tích được: x(1 – x) = 0

– suy ra : x = 0 hoặc : 1 – x = 0 x = 2

– Vậy nghiệm của đa thức đã cho là x = 0; x = 2.

Cho các đa thức sau: P (x) = −5x2 + 8x + 3; Q (x) = 2x2 − 9x + 6 Tính P(x) + Q(x) ; P(x) – Q(x) ; P(x) – Q(x)

\(\frac{7}{15}\) - \(\frac57\) + \(\frac{23}{15}\) + \(\frac57\) - \(\frac35\)

= (\(\frac{7}{15}\) + \(\frac{23}{15}\)) - (\(\frac57\) - \(\frac57\)) - \(\frac35\)

= 2 - 0 - \(\frac35\)

= 2 - \(\frac35\)

= \(\frac{10}{5}\) - \(\frac35\)

= \(\frac75\)

\(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)+\left(-\frac57+\frac57\right)-\frac35\)

\(=\frac{30}{15}-\frac35\)

\(=\frac{30}{15}-\frac{9}{15}\)

\(=\frac{21}{15}\)

✨ Bước 1: Rút gọn hai vế của phương trình

Vế phải:

\(8 , 5 - \frac{1}{2} = 8 , 0\)

Vậy phương trình trở thành:

\(2 \mid 5 - x \mid + \frac{1}{2} = 8\)

✨ Bước 2: Chuyển vế

Trừ \(\frac{1}{2}\) hai vế:

\(2 \mid 5 - x \mid = 8 - \frac{1}{2} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2}\)

✨ Bước 3: Chia hai vế cho 2

\(\mid 5 - x \mid = \frac{15}{4}\)

✨ Bước 4: Giải giá trị tuyệt đối

Ta có:

\(\mid 5 - x \mid = \frac{15}{4} \Rightarrow \left{\right. 5 - x = \frac{15}{4} \\ 5 - x = - \frac{15}{4}\)

Giải từng phương trình:

1. \(5 - x = \frac{15}{4} \Rightarrow x = 5 - \frac{15}{4} = \frac{20}{4} - \frac{15}{4} = \frac{5}{4}\)
2. \(5 - x = - \frac{15}{4} \Rightarrow x = 5 + \frac{15}{4} = \frac{20}{4} + \frac{15}{4} = \frac{35}{4}\)

✅ Kết luận:

Vậy phương trình có 2 nghiệm:

\(\boxed{x = \frac{5}{4} \text{ho}ặ\text{c} x = \frac{35}{4}}\)
Tk

7​251​−x​+​x−51​​+851​=1,2⇒​251​−x​+​x−51​​=1,2−851​⇒​251​−x​+​x−51​​=−7

Nhận xét:

\(\left{\right. \mid 2 \frac{1}{5} - x \mid \geq 0 , \forall x \\ \mid x - \frac{1}{5} \mid \geq 0 , \forall x \Rightarrow \mid 2 \frac{1}{5} - x \mid + \mid x - \frac{1}{5} \mid \geq 0 , \forall x\)
Mà \(- 7 < 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...

\(\frac{1}{2}x^2y.\left(\frac{-1}{2}x^3y\right)^3.\left(-2x^2\right)^2\)

\(=\frac{1}{2}.\left(-\frac{1}{8}\right).4.x^2y.x^9.y^3.x^4\)

\(=-\frac{1}{4}x^{15}y^4\)

Với \(x=2,y=-1\) ta có :

\(-\frac{1}{4}.2^{15}.\left(-1\right)^4=-2^{13}\)

Cảm ơn bạn nhé!

Sửa đề: \(\left(\frac35\right)^{10}\cdot\left(\frac53\right)^{10}-\frac{13^4}{39^4}+2024^0\)

\(=\left(\frac35\cdot\frac53\right)^{10}-\frac{13^4}{13^4\cdot3^4}+1\)

\(=1^{10}-\frac{1}{3^4}+1=2-\frac{1}{81}=\frac{161}{81}\)

Bài 3:

a: \(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\cdots\left|x+\frac{1}{2019\cdot2020}\right|=2020x\) (1)

=>2020x>=0

=>x>=0

Phương trình (1) sẽ trở thành:

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{2019\cdot2020}=2020x\)

=>\(2020x=2019x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\right)\)

=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2019}-\frac{1}{2020}\)

=>\(x=1-\frac{1}{2020}=\frac{2019}{2020}\)

b: \(\left|x+\frac{1}{1\cdot3}\right|+\left|x+\frac{1}{3\cdot5}\right|+\cdots+\left|x+\frac{1}{197\cdot199}\right|=100x\) (2)

=>100x>=0

=>x>=0

(2) sẽ trở thành: \(x+\frac{1}{1\cdot3}+x+\frac{1}{3\cdot5}+\cdots+x+\frac{1}{197\cdot199}=100x\)

=>\(100x=99x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{197\cdot199}\right)\)

=>\(x=\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{197}-\frac{1}{199}\right)=\frac12\left(1-\frac{1}{199}\right)\)

=>\(x=\frac12\cdot\frac{198}{199}=\frac{99}{199}\)

c: \(\left|x+\frac12\right|+\left|x+\frac16\right|+\left|x+\frac{1}{12}\right|+\cdots+\left|x+\frac{1}{110}\right|=11x\left(3\right)\)

=>11x>=0

=>x>=0

(3) sẽ trở thành:

\(11x=x+\frac12+x+\frac16+\ldots+x+\frac{1}{110}\)

=>\(11x=10x+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)

=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\) (nhận)

Bài 2:

a: \(\left|5-\frac23x\right|\ge0\forall x;\left|\frac23y-4\right|\ge0\forall y\)

Do đó: \(\left|5-\frac23x\right|+\left|\frac23y-4\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}5-\frac23x=0\\ \frac23y-4=0\end{cases}\Rightarrow\begin{cases}\frac23x=5\\ \frac23y=4\end{cases}\Rightarrow\begin{cases}x=5:\frac23=\frac{15}{2}\\ y=4:\frac23=6\end{cases}\)

b: \(\left|\frac23-\frac12+\frac34x\right|=\left|\frac34x+\frac16\right|\ge0\forall x\)

\(\left|1,5-\frac34-\frac32y\right|=\left|\frac34-\frac32y\right|\ge0\forall y\)

Do đó: \(\left|\frac34x+\frac16\right|+\left|\frac34-\frac32y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}\frac34x+\frac16=0\\ \frac34-\frac32y=0\end{cases}\Rightarrow\begin{cases}\frac34x=-\frac16\\ \frac32y=\frac34\end{cases}\Rightarrow\begin{cases}x=-\frac16:\frac34=-\frac16\cdot\frac43=-\frac{4}{18}=-\frac29\\ y=\frac34:\frac32=\frac24=\frac12\end{cases}\)

c: \(\left|x-2020\right|\ge0\forall x;\left|y-2021\right|\ge0\forall y\)

Do đó: \(\left|x-2020\right|+\left|y-2021\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2020=0\\ y-2021=0\end{cases}\Rightarrow\begin{cases}x=2020\\ y=2021\end{cases}\)

d: \(\left|x-y\right|\ge0\forall x,y\)

\(\left|y+\frac{21}{10}\right|\ge0\forall y\)

Do đó: \(\left|x-y\right|+\left|y+\frac{21}{10}\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ y+\frac{21}{10}=0\end{cases}\Rightarrow x=y=-\frac{21}{10}\)

Bài 1:

a: \(\left|\frac32x+\frac12\right|=\left|4x-1\right|\)

=>\(\left[\begin{array}{l}4x-1=\frac32x+\frac12\\ 4x-1=-\frac32x-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}4x-\frac32x=\frac12+1\\ 4x+\frac32x=-\frac12+1\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac52x=\frac32\\ \frac{11}{2}x=\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32:\frac52=\frac35\\ x=\frac12:\frac{11}{2}=\frac{1}{11}\end{array}\right.\)

b: \(\left|\frac75x+\frac12\right|=\left|\frac43x-\frac14\right|\)

=>\(\left[\begin{array}{l}\frac75x+\frac12=\frac43x-\frac14\\ \frac75x+\frac12=\frac14-\frac43x\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac75x-\frac43x=-\frac14-\frac12\\ \frac75x+\frac43x=\frac14-\frac12\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac{1}{15}x=-\frac34\\ \frac{41}{15}x=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac34:\frac{1}{15}=-\frac34\cdot15=-\frac{45}{4}\\ x=-\frac14:\frac{41}{15}=-\frac14\cdot\frac{15}{41}=-\frac{15}{164}\end{array}\right.\)

c: \(\left|\frac54x-\frac72\right|-\left|\frac58x+\frac35\right|=0\)

=>\(\left|\frac54x-\frac72\right|=\left|\frac58x+\frac35\right|\)

=>\(\left[\begin{array}{l}\frac54x-\frac72=\frac58x+\frac35\\ \frac54x-\frac72=-\frac58x-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x-\frac58x=\frac35+\frac72\\ \frac54x+\frac58x=-\frac35+\frac72\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac58x=\frac{41}{10}\\ \frac{15}{8}x=\frac{29}{10}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{41}{10}:\frac58=\frac{41}{10}\cdot\frac85=\frac{164}{25}\\ x=\frac{29}{10}:\frac{15}{8}=\frac{29}{10}\cdot\frac{8}{15}=\frac{116}{75}\end{array}\right.\)

d: \(\left|\frac78x+\frac56\right|-\left|\frac12x+5\right|=0\)

=>\(\left|\frac78x+\frac56\right|=\left|\frac12x+5\right|\)

=>\(\left[\begin{array}{l}\frac78x+\frac56=\frac12x+5\\ \frac78x+\frac56=-\frac12x-5\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac78x-\frac12x=5-\frac56\\ \frac78x+\frac12x=-5-\frac56\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac38x=\frac{25}{6}\\ \frac{11}{8}x=-\frac{35}{6}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{25}{6}:\frac38=\frac{25}{6}\cdot\frac83=\frac{200}{18}=\frac{100}{9}\\ x=-\frac{35}{6}:\frac{11}{8}=-\frac{35}{6}\cdot\frac{8}{11}=-\frac{140}{33}\end{array}\right.\)

lI dau la lI

\(B=(-\dfrac{1}{2};3]\)

\(B=(-\frac12;3]\)

\(\left(\dfrac{3}{4}+\dfrac{1}{2}\right)\times\dfrac{4}{5}:2\)

\(=\left(\dfrac{3}{4}+\dfrac{2}{4}\right)\times\dfrac{4}{5}\times\dfrac{1}{2}\)

\(=\dfrac{5}{4}\times\dfrac{4}{5}\times\dfrac{1}{2}\)

\(=1\times\dfrac{1}{2}=\dfrac{1}{2}\)

\(\left(\dfrac{3}{4}+\dfrac{1}{2}\right)\times\dfrac{4}{5}:2\\ =\left(\dfrac{3}{4}+\dfrac{2}{4}\right)\times\dfrac{4}{5}\times\dfrac{1}{2}\\ =\dfrac{5}{4}\times\dfrac{4}{5}\times\dfrac{1}{2}\\ =\dfrac{1}{2}\)

a: \(\left(x-\frac12\right)^2=0\)

=>\(x-\frac12=0\)

=>\(x=\frac12\)

b: \(\left(x-2\right)^2=1\)

=>\(\left[\begin{array}{l}x-2=1\\ x-2=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1+2=3\\ x=-1+2=1\end{array}\right.\)

c: \(\left(2x-1\right)^3=-8\)

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

=>2x-1=-2

=>2x=-1

=>\(x=-\frac12\)