2|x-3|+|2x+5|=11
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H9
HT.Phong (9A5)
CTVHS
30 tháng 8 2023
e) \(\left(x+3\right)^3=\left(2x\right)^3\)
\(\Rightarrow x+3=2x\)
\(\Rightarrow2x-x=3\)
\(\Rightarrow x=3\)
f) \(\left(5-x\right)^5=32\)
\(\Rightarrow\left(5-x\right)^5=2^5\)
\(\Rightarrow5-x=2\)
\(\Rightarrow x=5-2\)
\(\Rightarrow x=3\)
g) \(\left(5x-6\right)^3=64\)
\(\Rightarrow\left(5x-6\right)^3=4^3\)
\(\Rightarrow5x-6=4\)
\(\Rightarrow5x=4+6\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=\dfrac{10}{2}\)
\(\Rightarrow x=5\)
h) \(5\cdot9^x=405\)
\(\Rightarrow9^x=\dfrac{405}{5}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
i) \(11^5:11^{n-2}=11^5\)
\(\Rightarrow11^{n-2}=11^5:11^5\)
\(\Rightarrow11^{n-2}=1\)
\(\Rightarrow11^{n-2}=11^0\)
\(\Rightarrow n-2=0\)
\(\Rightarrow n=2\)
k) \(\left(3x\right)^3=\left(2x+1\right)^3\)
\(\Rightarrow3x=2x+1\)
\(\Rightarrow3x-2x=1\)
\(\Rightarrow x=1\)
P
1
2 tháng 2 2024
b: 2x+5=x-5
=>2x-x=-5-5
=>x=-10
c: 2x(x+2)+5(x-2)=0
=>\(2x^2+4x+5x-10=0\)
=>\(2x^2+9x-10=0\)
\(\text{Δ}=9^2-4\cdot2\cdot\left(-10\right)=81+80=161>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-9-\sqrt{161}}{4}\\x_2=\dfrac{-9+\sqrt{161}}{4}\end{matrix}\right.\)
h:
ĐKXĐ: \(x\notin\left\{2;-1\right\}\)
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
=>\(2\left(x-2\right)-\left(x+1\right)=3x-11\)
=>2x-4-x-1=3x-11
=>x-5=3x-11
=>x-3x=-11+5
=>-2x=-6
=>x=3(nhận)
i: 3x-12=0
=>3x=12
=>x=12/3=4
f: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)
=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)
=>\(\dfrac{3x-9+10x+5}{15}=6\)
=>13x-4=90
=>13x=94
=>\(x=\dfrac{94}{13}\)
24 tháng 10 2021
b: \(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)
\(=2x^2-6x-2x^2+4x+2x-4\)
=-4
11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
NK
2
7 tháng 8 2021
1.|x+1|+|3−2x|=|3x−2|
mà |3−2x|=|3x−2| nên |x+1|=0 => x+1=0 =>x=-1
HP
15 tháng 9 2021
a, ĐK: \(x\ge11\)
\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)
\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)
\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)
\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)
Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{}8\)
\(\Rightarrow\) phương trình vô nghiệm.
15 tháng 9 2021
\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)
\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)
Tìm x nha
Ta có:\(2\left|x-3\right|+\left|2x+5\right|=\left|6-2x\right|+\left|2x+5\right|\ge\left|\left(6-2x\right)+\left(2x+5\right)\right|=11,\forall x\)
\(Do\text{đ}\text{ó}2\left|x-3\right|+\left|2x+5\right|=11\Rightarrow\left(6-2x\right)\left(2x+5\right)\ge0\Rightarrow\frac{-5}{2}\le x\le3\)