Đại số lớp 8

c) x( 2x - 3 ) - 2( 3 - 2x) =0

\(\Leftrightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=\frac{3}{2}\end{array}\right.\)

d) 25x² - 36 =0

\(\Leftrightarrow\left(5x\right)^2-6^2=0\)

\(\Leftrightarrow\left(5x-6\right)\left(5x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-6=0\\5x+6=0\end{array}\right.\)

\(\Leftrightarrow x=\pm\frac{6}{5}\)

a) \(x\left(2x-3\right)-2\left(3-2x\right)=0\)

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

=>\(\left[\begin{array}{nghiempt}2x-3=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-2\end{array}\right.\)

b) \(25x^2-36=0\)

\(\Leftrightarrow\left(5x-6\right)\left(5x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-6=0\\5x+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{6}{5}\\x=-\frac{6}{5}\end{array}\right.\)

bằng 114,75

3(x-3)(x+7)+(x-4)^2+48=3x^2+12x-63+x^2-8x+63

=4x^2+4x=4x(x+1)

nhớ ticks nha

Ta có: 2n²+5n-1

=(2n²+2n+2n)+n-1

=2n​(n+2)+n-1

=(2n-1)(2n+2)

Vì 2n-1chia hết cho 2n-1 nên suy ra (2n-1)(2n+2) chia hết cho 2n-1​

Vậy 2n²+5n-1 chia hết cho 2n-1

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)

\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)

\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)

\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)

\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)

\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)

\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)

\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)

Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)

Ta có :

\(\begin{cases}\left(x+3y-6\right)^{2006}\ge0\\\left|2x-y-5\right|\ge0\end{cases}\)

\(\Rightarrow\begin{cases}x+3y-6=0\\2x-y-5=0\end{cases}\)

\(\Rightarrow\begin{cases}x=6-3y\\2x=y+5\end{cases}\)

\(\Rightarrow\begin{cases}2x=12-6y\\2x=y+5\end{cases}\)

\(\Rightarrow12-6y=y+5\)

\(\Rightarrow y=1\)

\(\Rightarrow x=6\)

-> x + y = 7

\(\left(x+1\right)\left(x-2\right)^2+x^2\left(4-x\right)=13\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)+4x^2-x^3=13\)

\(\Leftrightarrow x^3-4x^2+4x+x^2-4x+4+4x^2-x^3=13\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\)

Vậy x={-3;3}

0;1

nhầm cho xl

Ta có: \(\left|x\right|\le4\)

\(\Rightarrow x\le\pm4\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

Vậy \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

x ϵ {1,-1,2,-3,3,-2,4,-4,}

\(\left(x-2\right)\left(x^2+2x+4\right)+35=0\)

=> \(x^3-8+35=0\)

=> \(x^3=-27\)

=> \(x=-3\)

Ta có :

\(\left(x-2\right)\left(x^2+2x+4\right)+35=0\)

\(\Rightarrow x^3-2^3+35=8\)

\(\Rightarrow x^3=-27\)

=> x = - 3

Vậy x = - 3