a)

\(\frac{x}{2}=\frac{y}{4}\)

\(\Rightarrow\frac{x^4}{16}=\frac{y^4}{256}=\frac{x^2y^2}{2^2.4^2}=\frac{4}{64}=\frac{1}{16}\)

\(\Rightarrow\begin{cases}x=\pm1\\y=\pm2\end{cases}\)

Mà 2 ; 4 cùng dấu

=> x ; y cùng dấu

Vậy ........

b)

\(4x=7y\)

\(\Rightarrow\frac{x}{7}=\frac{y}{4}\)

\(\Rightarrow\frac{x^2}{49}=\frac{y^2}{16}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có :

\(\frac{x^2}{49}=\frac{y^2}{16}=\frac{x^2+y^2}{49+16}=\frac{260}{65}=4\)

\(\Rightarrow\begin{cases}x=\pm14\\y=\pm8\end{cases}\)

Mày 4 và 7 cùng dấu

=> x ; y cùng dấu

Vậy ........

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

a)

\(\frac{16}{2^x}=2\)

\(\Rightarrow2^{x+1}=16\)

\(\Rightarrow2^{x+1}=2^4\)

\(\Rightarrow x+1=4\)

\(\Rightarrow x=3\)

b)

\(\frac{\left(-3\right)^x}{81}=-27\)

\(\Rightarrow\left(-3\right)^x=-\left(3^3.3^4\right)\)

\(\Rightarrow-3^x=-3^7\)

=> x=7

c)

\(8^n:2^n=4\)

\(\Rightarrow2^{3n}:2^n=4\)

\(\Rightarrow2^{3n-n}=4\)

\(\Rightarrow2^{2n}=2^2\)

=>2n=2

=>n=1

a)\(\frac{16}{2^n}=2\)

=>16:2ⁿ=2

=>2ⁿ=16:2

=>2ⁿ=8

b)ko nhớ cách làm

c)8ⁿ:2ⁿ=4

=>(2³)ⁿ:2ⁿ=2²

=>2³ⁿ:2ⁿ=2²

=>2^3n-n=2²

=>2²ⁿ=2²

=>2n=2

=>n=1

dc rùi chứ

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+6-4}=\frac{2x-2+3y-6-z+3}{4+6-4}\)

\(=\frac{\left(2x+3y-z\right)+\left(-2+6+3\right)}{6}=\frac{50+\left(-5\right)}{6}=\frac{45}{6}=7,5\)

\(\frac{x-1}{2}=7,5\Rightarrow x-1=15\Rightarrow x=16\)

\(\frac{y-2}{3}=7,5\Rightarrow y-2=24,5\Rightarrow y=20,5\)

\(\frac{z-3}{4}=7,5\Rightarrow z-3=30\Rightarrow z=33\)

a) \(\begin{cases}\left(x+2\right)^2\ge0\\\left(y-\frac{1}{5}\right)^2\ge0\end{cases}\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10=-10\)hay \(C\ge-10\)

Dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-\frac{1}{5}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+2=0\\y-\frac{1}{5}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}}}\)

Vậy GTNN C là -10 khi \(\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}.}\)

b)\(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+5\ge0+5=5\)

\(\Rightarrow\frac{4}{\left(2x-3\right)^2-5}\le\frac{4}{5}\Leftrightarrow D\le\frac{4}{5}\)

Dấu "=" xảy ra khi:

\(\left(2x-3\right)^2=0\Rightarrow2x-3=0\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)

Vậy GTLN D là \(\frac{4}{5}\)khi \(x=\frac{3}{2}.\)

thank bạn nha

\(f\left(x\right)=4x^2+3x+1\)

\(g\left(x\right)=3x^2-2x+1.\)

a) \(h\left(x\right)=f\left(x\right)-g\left(x\right)\)

\(\Rightarrow h\left(x\right)=\left(4x^2+3x+1\right)-\left(3x^2-2x+1\right)\)

\(\Rightarrow h\left(x\right)=4x^2+3x+1-3x^2+2x-1\)

\(\Rightarrow h\left(x\right)=\left(4x^2-3x^2\right)+\left(3x+2x\right)+\left(1-1\right)\)

\(\Rightarrow h\left(x\right)=x^2+5x.\)

b) Ta có \(h\left(x\right)=x^2+5x.\)

Đặt \(x^2+5x=0\)

\(\Rightarrow x.\left(x+5\right)=0\)

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

Vậy \(x=0\) và \(x=-5\) là các nghiệm của đa thức \(h\left(x\right).\)

Chúc bạn học tốt!

mơn nhé