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

=>\(1+x=2x\)

=>\(x-2x=-1\)

=>\(-x=-1\)

=>\(x=1\)

vậy ​\(x=1\)​

b,\(\left(x-1\right)^2=16\)

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

=>\(x-1=4\)

=>\(x=4+1\)

=>\(x=5\)

Vậy​\(x=5\)​

c,\(\left(x+1\right)^2=25\)

=>\(\left(x+1\right)^2=5^2\)

=>\(x+1=5\)

=>\(x=5-1\)

=>\(x=4\)

Vậy \(x=4\)

d,\(4x^3+15=47\)

=>\(4x^3=47-15\)

=>\(4x^3=32\)

=>\(x^3=32:4\)

=>\(x^3=8\)

=>\(x^3=2^3\)

=>\(x=2\)

Vậy\(x=2\)

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

=>\(2x-1=x\)

=>\(2x-x=1\)

=>\(x=1\)

Vậy\(x=1\)

ĐÚNG K MÌNH NHA

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

2) \(\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)

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

5) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x\right)^3-\left(3y\right)^3=8x^3-27y^3\)

6) \(\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)

3) Đề thiếu

\(1,\left(x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)\left(x-1\right)^2\)
\(2,\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)
\(3,\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+2\)
\(4,\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\left(\frac{1}{2}\right)^3-x^3=\frac{1}{8}-x^3\)
\(5,\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x\right)^3-\left(3y\right)^3=8x^3-27y^3\)
\(6,\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)

a,x²-25-(x+5)                                                   b,mình quên mất rồi.Đợi tí nhé

(x²-25)-(x+5)=0

(x²-5²)+(x-5)=0

(x-5)(x+5)+(x-5)=0

(x-5)(x+5+1)=0

x-5=0 hoặc x+5+1=0

x=0+5 hoặc x=0-5-1

x=5     hoặc x=-6

Vậy x=5 và x=-6

Giải bpt

A)  (x^2+1)×(4x-2)≫0(lớn hơn hoặc =0)

B) (x-2)×x^2>0

Mog mn giúp ạ

E cần gấp

Thak mn

\(x+2x+3x+4x+10x+50=100\)

\(\Leftrightarrow x\left(1+2+3+4+10\right)+50=100\)

\(\Leftrightarrow x20=50\Leftrightarrow x=\frac{5}{2}\)

x+2x+3x+4x+10x+50=20x+50=100

=> 20x=50

=> \(x=\frac{5}{2}\)

Vậy x=\(\frac{5}{2}\)

                                     Bài làm :

Ta có :

\(x+2x+3x+4x+10x+50=100\)

\(\Leftrightarrow20x=100-50=50\)

\(\Leftrightarrow x=\frac{50}{20}=\frac{5}{2}=2,5\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

k ho cai

k hộ please

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Câu 1 :

a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

b. 

|-2x|=5x+14

 Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14

 <=> - 2x = 5x + 14

 <=> - 2x - 5x = 14

 <=> - 7x = 14

 <=> x = - 2 (thoã mãn)

 Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.

Ta có pt: 2x = 5x + 14

 <=> - 3x = 14

<=> x = \(-\dfrac{14}{3}\)
 Vậy pt có nghiệm x = - 2

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)

\(a,4x-5=23\)

\(\Leftrightarrow4x=23+5\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(b,\left|-2x\right|=5x+14\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)

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

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

\(\Leftrightarrow x=0\)

Vậy \(S=\left\{0\right\}\)

a) \(4x-5=23\)

    \(4x=23+5\)

      \(4x=28\)

        \(x=7\)

b) \(\left|-2x\right|=5x+14\)

 \(\Leftrightarrow\)   \(-2x-5=14\)

\(\Leftrightarrow\)    \(-7x=14\)

\(\Leftrightarrow\)         \(x=-2\)

\(\Leftrightarrow\)    \(-2x=-\left(5x+14\right)\)

\(\Leftrightarrow\)    \(-2x=-\left(5x-14\right)\)

\(\Leftrightarrow\)  \(-2x+5x=-14\)

 \(\Leftrightarrow\)    \(3x=-14\)

 \(\Leftrightarrow\) \(x=-\dfrac{14}{3}\) \(\left(\text{vô lí}\right)\)

  \(\Leftrightarrow x=-2\)   

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)

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

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

\(\Leftrightarrow x^2+x+2=x^2+2\)

\(\Leftrightarrow x+2=2\)

\(\Leftrightarrow x=0\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3