Bài 2: Phương trình bậc nhất một ẩn và cách giải

\(\left(x^2+5x+4\right)\left(x^2-4x+4\right)=10x^2\)

x= 0 không phải nghiệm

chia hai vế cho x^4

\(\left(x+\dfrac{4}{x}+5\right)\left(x+\dfrac{4}{x}-4\right)=10\)

Đặt x+4/x =t

\(t^2+t-20=10\)\(\Rightarrow\left[{}\begin{matrix}t=5\\t=-6\end{matrix}\right.\)

Thay lại tìm x tự làm

a) (x²-5x+x-5)-x²+6x-3x-7=0

<=> -2x-12=0

<=> x=-6

vậy S=(-6)

b) \(\dfrac{x\left(x-2\right)-\left(2x-1\right)\left(x+3\right)}{x\left(x+3\right)}=\dfrac{11-2x^2}{x\left(x+3\right)}\)

Đk: x \(\ne\) 0,-3

=> \(x^2-2x-\left(2x^2+6x-x-3\right)=11-2x^2\)

<=> \(x^2-2x-2x^2-6x+x+3-11+2x^2=0\)

\(x^2-7x-8=0\)

\(\left(x-8\right)\left(x+1\right)=0\)

=> x=8 hoặc x=-1

a) \(\left(x+1\right)\left(x-5\right)-x\left(x-6\right)=3x+7\)

\(\Leftrightarrow x^2+x-5x-5-x^2+6x=3x+7\)

\(\Leftrightarrow x^2-x^2+x-5x+6x-3x=7+5\)

\(\Leftrightarrow-x=12\)

\(\Leftrightarrow x=-12\)

Vậy phương trình có tập nghiệm \(S=\left\{-12\right\}\)

b) \(\dfrac{x-2}{x+3}-\dfrac{2x-1}{x}=\dfrac{11-2x^2}{x^2+3x}\)

\(\Leftrightarrow\dfrac{\left(x-2\right)x}{\left(x+3\right)x}-\dfrac{\left(2x-1\right)\left(x+3\right)}{x\left(x+3\right)}=\dfrac{11-2x^2}{x^2+3x}\)

\(\Leftrightarrow\dfrac{x^2-2x}{x^2+3x}-\dfrac{2x^2-x+6x-3}{x^2+3x}=\dfrac{11-2x^2}{x^2+3x}\)

\(\Leftrightarrow x^2-2x-2x^2-5x+3=11-2x^2\)

\(\Leftrightarrow x^2-2x^2+2x^2-2x-5x+3-11=0\)

\(\Leftrightarrow x^2-7x-8=0\)

\(\Leftrightarrow x^2-8x+x-8=0\)

\(\Leftrightarrow x\left(x-8\right)+\left(x-8\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1\\8\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{-1;8\right\}\)

đặt \(f\left(x\right)=\left(2x-1\right)\left(3-2x\right)\left(1-x\right)\)

giải \(f\left(x\right)\)=0\(\Rightarrow x=\dfrac{1}{2},x=\dfrac{3}{2},x=1\)

Bảng xét dấu:

ta có: f(x)<0\(\Rightarrow\)1<x<\(\dfrac{3}{2}\)

cái chỗ \(+\infty\) đến \(\dfrac{1}{2}\) là âm nha bn

vậy là thêm trường hợp x<\(\dfrac{1}{2}\)

Đặt A=\(\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)

Gọi \(\dfrac{a-b}{c}=x\); \(\dfrac{b-c}{a}=y\); \(\dfrac{c-a}{b}=z\) => \(\dfrac{c}{a-b}=\dfrac{1}{x};\dfrac{a}{b-c}=\dfrac{1}{y};\dfrac{b}{c-a}=\dfrac{1}{z}\)

=> A=(x+y+z)\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(=1+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+1+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}+1\)

= \(\dfrac{x+z}{y}+\dfrac{x+y}{z}+\dfrac{z+y}{x}+3\)

Lại có: \(\dfrac{z+y}{x}=\dfrac{\dfrac{b-c}{a}+\dfrac{c-a}{b}}{\dfrac{a-b}{c}}\) = \(\dfrac{b^2-bc+ac-a^2}{ab}.\dfrac{c}{a-b}\)= \(\dfrac{\left(b-a\right)\left(a+b\right)-c\left(b-a\right)}{ab}.\dfrac{c}{a-b}\) =\(\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}.\dfrac{c}{a-b}\) = \(\dfrac{-\left(a-b\right)\left(a+b-c\right)}{ab}.\dfrac{c}{a-b}\)= \(\dfrac{\left(-a-b+c\right).c}{ab}\) ⁽¹⁾

Lại có: a+b+c=0 <=> c=-a-b

Thay vào (1) ta được:\(\dfrac{z+y}{x}\)= \(\dfrac{2c^2}{ab}\)

Tương tự ta chứng minh được: \(\dfrac{x+z}{y}=\dfrac{2a^2}{bc}\) ; \(\dfrac{x+y}{z}=\dfrac{2b^2}{ac}\)

=> A=\(\dfrac{2a^2}{bc}+\dfrac{2b^2}{ac}+\dfrac{2c^2}{ab}\)+3 = \(\dfrac{2a^3}{abc}+\dfrac{2b^3}{abc}+\dfrac{2c^3}{abc}+3=\dfrac{2\left(a^3+b^3+c^3\right)}{abc}+3\)

ta chứng minh được \(a^3+b^3+c^3=3abc\) khi a+b+c=0

=> \(A=\dfrac{2.3abc}{abc}+3=6+3=9\)

Vậy A=9

<=> \(\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)=9

các bạn giúp mình câu này với

\(18+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-36\right]\)

\(\Leftrightarrow\dfrac{198}{11}+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-\dfrac{396}{11}\right]\)

\(\Leftrightarrow\dfrac{198+x-18}{11}=36+\dfrac{1}{11}\times\dfrac{10x-180-396}{11}\)

\(\Leftrightarrow\dfrac{180+x}{11}=36+\dfrac{10x-576}{121}\)

\(\Leftrightarrow\dfrac{1980+11x}{121}=\dfrac{4356}{121}+\dfrac{10x-576}{121}\)

\(\Leftrightarrow1980+11x=4356+10x-576\)

\(\Leftrightarrow11x-10x=4356-1980-576\)

\(\Leftrightarrow x=1800\)

\(x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}=0\)

\(\Rightarrow\left(x^6-x^5\right)+\left(x^4-x^3\right)+\left(x^2-x\right)+\dfrac{3}{4}=0\)

\(\Rightarrow x^5\left(x-1\right)+x^3\left(x-1\right)+x\left(x-1\right)+\dfrac{3}{4}=0\)

\(\Rightarrow\left(x-1\right)\left(x^5+x^3+x\right)+\dfrac{3}{4}=0\)

.... bí cmnr :))

a, 2x-3>5

<=> 2x>5+3

<=> 2x>8

<=> x>4

Vậy S={x|x>4}

a) \(2x-3>5\)

\(\Leftrightarrow2x>8\)

\(\Leftrightarrow x>4\)

Tập nghiệm: \(S=\left\{x|x>4\right\}\)

b) \(\dfrac{x-1}{2}-1\le\dfrac{2\left(x+1\right)}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)}{6}-\dfrac{6}{6}\le\dfrac{4\left(x+1\right)}{6}\)

\(\Leftrightarrow3x-3-6\le4x+4\)

\(\Leftrightarrow-x\le13\)

\(\Leftrightarrow x\ge-13\)

Tập nghiệm: \(S=\left\{x|x\ge-13\right\}\)

\(\dfrac{2012}{x}< 2013\\ \Leftrightarrow2012< 2013x\\ \Leftrightarrow\dfrac{2012}{2013}< x\)

vậy bất phương trình có tập nghiệm là \(S=\left\{x|x>\dfrac{2012}{2013}\right\}\)

Ta co

\(\dfrac{2012}{x}< 2013\)

\(\Leftrightarrow\dfrac{2012}{x}< \dfrac{2013x}{x}\)

\(\Leftrightarrow\)2012<2013x

\(\Leftrightarrow\)-x\(< -\dfrac{2012}{2013}\)

\(\Leftrightarrow\)x<\(\dfrac{2012}{2013}\)

Vay tap nghiem cua bat phuong trinh la \(\left\{x|x< \dfrac{2012}{2013}\right\}\)

\(\dfrac{2012}{x}< 2013\)

\(\Leftrightarrow2012< 2013x\)

\(\Leftrightarrow\dfrac{2012}{2013}< x\)

Vậy bất phương trình có tập nghiệm: \(S=\left\{x|x>\dfrac{2012}{2013}\right\}\)